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8.17 Complete each synthesis by giving missing starting material, reagent or products

8.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO (iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO CHO (v) (vi) PhCOPh

8.2 Name the following compounds according to IUPAC system of nomenclature: (i) CH3CH(CH3)CH2CH2CHO (ii) CH3CH2COCH(C2H5)CH2CH2Cl (iii) CH3CH=CHCHO (iv) CH3COCH2COCH3 (v) CH3CH(CH3)CH2C(CH3)2COCH3 (vi) (CH3)3CCH2COOH (vii) OHCC6H4CHO-p

8.6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then H3O + (ii) Tollens’ reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid 8.7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal 8.8 How will you convert ethanal into the following compounds? (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid 8.9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile. 8.10 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. 8.11 An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved. 8.12 Arrange the following compounds in increasing order of their property as indicated: (i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) 8.13 Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal 8.14 How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom (i) Methyl benzoate (ii) m-Nitrobenzoic acid (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde. 8.15 How will you bring about the following conversions in not more than two steps? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to a-Hydroxyphenylacetic acid (ix) Benzoic acid to m- Nitrobenzyl alcohol 8.16 Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation Chemistry 256 Reprint 2025-26

8.3 STRUCTURAL RepresenTATIONS Similarly, CH3CH2CH2CH2CH2CH2CH2CH3 OF organic COMPOUNDs can be further condensed to CH3(CH2)6CH3. 8.3.1 Complete, Condensed and Bond-line For further simplification, organic chemists Structural Formulas use another way of representing the structures, in which only lines are used.Structures of organic compounds are In this bond-line structural representationrepresented in several ways. The Lewis of organic compounds, carbon andstructure or dot structure, dash structure, hydrogen atoms are not shown and thecondensed structure and bond line structural lines representing carbon-carbon bonds areformulas are some of the specific types. The drawn in a zig-zag fashion. The only atomsLewis structures, however, can be simplified specifically written are oxygen, chlorine,by representing the two-electron covalent nitrogen etc. The terminals denote methylbond by a dash (–). Such a structural formula (–CH3) groups (unless indicated otherwise byfocuses on the electrons involved in bond a functional group), while the line junctionsformation. A single dash represents a single denote carbon atoms bonded to appropriatebond, double dash is used for double bond number of hydrogens required to satisfy theand a triple dash represents triple bond. Lone- valency of the carbon atoms. Some of thepairs of electrons on heteroatoms (e.g., oxygen, examples are represented as follows:nitrogen, sulphur, halogens etc.) may or may (i) 3-Methyloctane can be represented innot be shown. Thus, ethane (C2H6), ethene various forms as:(C2H4), ethyne (C2H2) and methanol (CH3OH) can be represented by the following structural (a) CH3CH2CHCH2CH2CH2CH2CH3 formulas. Such structural representations are | called complete structural formulas. CH3 (b) Ethane Ethene (c) Ethyne Methanol These structural formulas can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by (ii) Various ways of representing 2-bromo indicating the number of identical groups butane are: attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula. Thus, (a) CH3CHBrCH2CH3 (b)ethane, ethene, ethyne and methanol can be written as: CH3CH3 H2C=CH2 HC≡CH CH3OH (c) Ethane Ethene Ethyne Methanol Reprint 2025-26 organic chemistry – some basic principles and techniques 259 In cyclic compounds, the bond-line formulas may be given as follows: (b) Solution Condensed formula: Cyclopropane (a) HO(CH2)3CH(CH3)CH(CH3)2 (b) HOCH(CN)2 Bond-line formula: (a) Cyclopentane (b) chlorocyclohexane Problem 8.6 Problem 8.4 Expand each of the following bond-line Expand each of the following condensed formulas to show all the atoms including formulas into their complete structural carbon and hydrogen formulas. (a) (a) CH3CH2COCH2CH3 (b) CH3CH=CH(CH2)3CH3 Solution (b) (a) (c) (b) (d) Solution Problem 8.5 For each of the following compounds, write a condensed formula and also their bond-line formula. (a) HOCH2CH2CH2CH(CH3)CH(CH3)CH3 Reprint 2025-26 260 chemistry Molecular Models Molecular models are physical devices that are used for a better visualisation and perception of three-dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available. Commonly three types of molecular models are used: (1) Framework model, (2) Ball-and-stick model, and (3) Space filling model. In the framework model only the bonds connecting the atoms of a molecule and not the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of atoms. In the ball-and-stick model, both the atoms and the bonds are shown. Balls represent atoms and the stick denotes a bond. Compounds containing C=C (e.g., ethene) can best be represented by using8.3.2 Three-Dimensional springs in place of sticks. These models are Representation of Organic referred to as ball-and-spring model. The Molecules space-filling model emphasises the relative The three-dimensional (3-D) structure of size of each atom based on its van der Waals organic molecules can be represented on radius. Bonds are not shown in this model. paper by using certain conventions. For It conveys the volume occupied by each atom in the molecule. In addition to these models,example, by using solid ( ) and dashed computer graphics can also be used for( ) wedge formula, the 3-D image of a molecular modelling. molecule from a two-dimensional picture can be perceived. In these formulas the solid-wedge is used to indicate a bond projecting out of the plane of paper, towards the observer. The dashed-wedge is used to depict the bond projecting out of the plane of the paper and away from the observer. Wedges are shown in such a way that the broad end of the wedge is towards the observer. The bonds Framework model Ball and stick model lying in plane of the paper are depicted by using a normal line (—). 3-D representation of methane molecule on paper has been shown in Fig. 8.1 Space filling model Fig. 8.2 Fig. 8.1 Wedge-and-dash representation of CH4 Reprint 2025-26 organic chemistry – some basic principles and techniques 261

8.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? Answers to Some Intext Questions 8.1 (i) (iv) (ii) (v) (iii) (vi) 257 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 8.2 (i) (ii) (iii) (iv) 8.3 CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH 8.4 (i) Butanone < Propanone < Propanal < Ethanal (ii) Acetophenone < p-Tolualdehyde , Benzaldehyde < p-Nitrobenzaldehyde. 8.5 (i) (ii) (iii) (iv) 8.6 (i) 3-Phenylpropanoic acid (ii) 3-Methylbut-2-enoic acid (iii) 2-Methylcyclopentanecarboxylic acid. (iv) 2,4,6-Trinitrobenzoic acid 8.7 (i) (ii) (iii) (iv) 8.8 Chemistry 258 Reprint 2025-26 UnitUnitUnitUnit Unit9 AminesAminesmines9minesminesminesminesminesminesminesObjectives After studying this Unit, you will be able to “The chief commercial use of amines is as intermediates in the · describe amines as derivatives of synthesis of medicines and fibres” . ammonia having a pyramidal structure; Amines constitute an important class of organic · classify amines as primary, compounds derived by replacing one or more hydrogen secondary and tertiary; atoms of ammonia molecule by alkyl/aryl group(s). In· name amines by common names nature, they occur among proteins, vitamins, alkaloids and IUPAC system; and hormones. Synthetic examples include polymers,· describe some of the important dye stuffs and drugs. Two biologically active methods of preparation of amines; compounds, namely adrenaline and ephedrine, both· explain the properties of amines; containing secondary amino group, are used to increase· distinguish between primary, blood pressure. Novocain, a synthetic amino compound, secondary and tertiary amines; is used as an anaesthetic in dentistry. Benadryl, a well· describe the method of prepara- known antihistaminic drug also contains tertiary amino tion of diazonium salts and their importance in the synthesis of a group. Quaternary ammonium salts are used as series of aromatic compounds surfactants. Diazonium salts are intermediates in the including azo dyes. preparation of a variety of aromatic compounds including dyes. In this Unit, you will learn about amines and diazonium salts. I. AMINES Amines can be considered as derivatives of ammonia, obtained by replacement of one, two or all the three hydrogen atoms by alkyl and/or aryl groups. For example: 9.19.19.19.19.1 StructureStructureStructureStructureStructure ofofofofof AminesAminesAminesAminesAmines Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp3 hybridised and the geometry of amines is pyramidal. Each of the three sp3 hybridised orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines. The fourth orbital of nitrogen in all amines contains an unshared pair of electrons. Due to the presence of unshared pair of electrons, the angle C–N–E, (where E is Reprint 2025-26 C or H) is less than 109.5°; for instance, it is 108o in case of trimethylamine as shown in Fig. 9.1. Fig. 9.1 Pyramidal shape of trimethylamine 9.29.29.29.29.2 ClassificationClassificationClassificationClassificationClassification Amines are classified as primary (1o), secondary (2o) and tertiary (3o) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule. If one hydrogen atom of ammonia is replaced by R or Ar , we get RNH2 or ArNH2, a primary amine (1o). If two hydrogen atoms of ammonia or one hydrogen atom of R-NH2 are replaced by another alkyl/aryl(R’) group, what would you get? You get R-NHR’, secondary amine. The second alkyl/aryl group may be same or different. Replacement of another hydrogen atom by alkyl/aryl group leads to the formation of tertiary amine. Amines are said to be ‘simple’ when all the alkyl or aryl groups are the same, and ‘mixed’ when they are different. 9.39.39.39.39.3 NomenclatureNomenclatureNomenclatureNomenclatureNomenclature In common system, an aliphatic amine is named by prefixing alkyl group to amine, i.e., alkylamine as one word (e.g., methylamine). In secondary and tertiary amines, when two or more groups are the same, the prefix di or tri is appended before the name of alkyl group. In IUPAC system, primary amines are named as alkanamines. The name is derived by replacement of ‘e’ of alkane by the word amine. For example, CH3NH2 is named as methanamine. In case, more than one amino group is present at different positions in the parent chain, their positions are specified by giving numbers to the carbon atoms bearing –NH2 groups and suitable prefix such as di, tri, etc. is attached to the amine. The letter ‘e’ of the suffix of the hydrocarbon part is retained. For example, H2N–CH2–CH2–NH2 is named as ethane-1, 2-diamine. To name secondary and tertiary amines, we use locant N to designate substituent attached to a nitrogen atom. For example, CH3 NHCH2CH3 is Chemistry 260 Reprint 2025-26 named as N-methylethanamine and (CH3CH2)3N is named as N, N- diethylethanamine. More examples are given in Table 9.1. In arylamines, –NH2 group is directly attached to the benzene ring. C6H5NH2 is the simplest example of arylamine. In common system, it is known as aniline. It is also an accepted IUPAC name. While naming arylamines according to IUPAC system, suffix ‘e’ of arene is replaced by ‘amine’. Thus in IUPAC system, C6H5–NH2 is named as benzenamine. Common and IUPAC names of some alkylamines and arylamines are given in Table 9.1. Table 9.1: Nomenclature of Some Alkylamines and Arylamines Amine Common name IUPAC name CH3-–CH2–NH2 Ethylamine Ethanamine CH3–CH2–CH2–NH2 n-Propylamine Propan-1-amine Isopropylamine Propan-2-amine Ethylmethylamine N-Methylethanamine Trimethylamine N,N-Dimethylmethanamine N,N-Diethylbutylamine N,N-Diethylbutan-1-amine Allylamine Prop-2-en-1-amine Hexamethylenediamine Hexane-1,6-diamine Aniline Aniline or Benzenamine o-Toluidine 2-Methylaniline p-Bromoaniline 4-Bromobenzenamine or 4-Bromoaniline N,N-Dimethylaniline N,N-Dimethylbenzenamine 261 Amines Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 9.1 Classify the following amines as primary, secondary or tertiary: 9.2 (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N. (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines? 9.49.49.49.49.4 PreparationPreparationPreparationPreparationPreparation Amines are prepared by the following methods: ofofofofof AminesAminesAminesAminesAmines 1. Reduction of nitro compounds Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium. Nitroalkanes can also be similarly reduced to the corresponding alkanamines. Reduction with iron scrap and hydrochloric acid is preferred because FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction. 2. Ammonolysis of alkyl halides You have read (Unit 6, Class XII) that the carbon - halogen bond in alkyl or benzyl halides can be easily cleaved by a nucleophile. Hence, an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group. This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis. The reaction is carried out in a sealed tube at 373 K. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt. Chemistry 262 Reprint 2025-26 The free amine can be obtained from the ammonium salt by treatment with a strong base: Ammonolysis has the disadvantage of yielding a mixture of primary, secondary and tertiary amines and also a quaternary ammonium salt. However, primary amine is obtained as a major product by taking large excess of ammonia. The order of reactivity of halides with amines is RI > RBr >RCl. Write chemical equations for the following reactions: ExampleExampleExampleExampleExample 9.19.19.19.19.1 (i) Reaction of ethanolic NH3 with C2H5Cl. (ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of CH3Cl. SolutionSolutionSolutionSolutionSolution 3. Reduction of nitriles Nitriles on reduction with lithium aluminium hydride (LiAlH4) or catalytic hydrogenation produce primary amines. This reaction is used for ascent of amine series, i.e., for preparation of amines containing one carbon atom more than the starting amine. 4. Reduction of amides The amides on reduction with lithium aluminium hydride yield amines. 263 Amines Reprint 2025-26 5. Gabriel phthalimide synthesis Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. 6. Hoffmann bromamide degradation reaction Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide. ExampleExampleExampleExampleExample 9.29.29.29.29.2 Write chemical equations for the following conversions: (i) CH3–CH2–Cl into CH3–CH2–CH2–NH2 (ii) C6H5–CH2–Cl into C6H5–CH2–CH2–NH2 SolutionSolutionSolutionSolutionSolution Chemistry 264 Reprint 2025-26 Write structures and IUPAC names of ExampleExampleExampleExampleExample 9.39.39.39.39.3 (i) the amide which gives propanamine by Hoffmann bromamide reaction. (ii) the amine produced by the Hoffmann degradation of benzamide. SolutionSolutionSolutionSolutionSolution (i) Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. Structure and IUPAC name of the starting amide with four carbon atoms are given below: Butanamide (ii) Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine containing six carbon atoms. Aniline or benzenamine IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 9.3 How will you convert (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl–(CH2)4–Cl into hexan-1,6-diamine? 9.59.59.59.59.5 PhysicalPhysicalPhysicalPhysicalPhysical The lower aliphatic amines are gases with fishy odour. Primary amines with three or more carbon atoms are liquid and still higher ones are PropertiesPropertiesPropertiesPropertiesProperties solid. Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation. Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water. Considering the electronegativity of nitrogen of amine and oxygen of alcohol as 3.0 and 3.5 respectively, you can predict the pattern of solubility of amines and alcohols in water. Out of butan-1-ol and butan-1-amine, which will be more soluble in water and why? Amines are soluble in organic solvents like alcohol, ether and benzene. You may remember that alcohols are more polar than amines and form stronger intermolecular hydrogen bonds than amines. Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows: 265 Amines Reprint 2025-26 Primary > Secondary > Tertiary Intermolecular hydrogen bonding in primary amines is shown in Fig. 9.2. Fig. 9.2 Intermolecular hydrogen bonding in primary amines Boiling points of amines, alcohols and alkanes of almost the same molar mass are shown in Table 9.2. Table 9.2: Comparison of Boiling Points of Amines, Alcohols and Alkanes of Similar Molecular Masses Sl. No. Compound Molar mass b.p./K 1. n-C4H9NH2 73 350.8 2. (C2H5)2NH 73 329.3 3. C2H5N(CH3)2 73 310.5 4. C2H5CH(CH3)2 72 300.8 5. n-C4H9OH 74 390.3 9.69.69.69.69.6 ChemicalChemicalChemicalChemicalChemical Difference in electronegativity between nitrogen and hydrogen atoms and the presence of unshared pair of electrons over the nitrogen atom makes ReactionsReactionsReactionsReactionsReactions amines reactive. The number of hydrogen atoms attached to nitrogen atom also decides the course of reaction of amines; that is why primary (–NH2), secondary N H and tertiary amines N differ in many reactions. Moreover, amines behave as nucleophiles due to the presence of unshared electron pair. Some of the reactions of amines are described below: 1. Basic character of amines Amines, being basic in nature, react with acids to form salts. Chemistry 266 Reprint 2025-26 Amine salts on treatment with a base like NaOH, regenerate the parent amine. Amine salts are soluble in water but insoluble in organic solvents like ether. This reaction is the basis for the separation of amines from the non basic organic compounds insoluble in water. The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base. Basic character of amines can be better understood in terms of their Kb and pKb values as explained below:       R  NH 3   OH  K =  R  NH2  H2 O        R  N H3   O H  or K [H 2 O] =  R  NH 2        R  N H 3   O H  or K b =  R  NH2  pKb = –log Kb Larger the value of Kb or smaller the value of pKb, stronger is the base. The pKb values of few amines are given in Table 9.3. pKb value of ammonia is 4.75. Aliphatic amines are stronger bases than ammonia due to +I effect of alkyl groups leading to high electron density on the nitrogen atom. Their pKb values lie in the range of 3 to 4.22. On the other hand, aromatic amines are weaker bases than ammonia due to the electron withdrawing nature of the aryl group. Table 9.3: pKb Values of Amines in Aqueous Phase Name of amine pKb Methanamine 3.38 N-Methylmethanamine 3.27 N,N-Dimethylmethanamine 4.22 Ethanamine 3.29 N-Ethylethanamine 3.00 N,N-Diethylethanamine 3.25 Benzenamine 9.38 Phenylmethanamine 4.70 N-Methylaniline 9.30 N,N-Dimethylaniline 8.92 267 Amines Reprint 2025-26 You may find some discrepancies while trying to interpret the Kb values of amines on the basis of +I or –I effect of the substituents present in amines. Besides inductive effect, there are other effects like solvation effect, steric hinderance, etc., which affect the basic strength of amines. Just ponder over. You may get the answer in the following paragraphs. Structure-basicity relationship of amines Basicity of amines is related to their structure. Basic character of an amine depends upon the ease of formation of the cation by accepting a proton from the acid. The more stable the cation is relative to the amine, more basic is the amine. (a) Alkanamines versus ammonia Let us consider the reaction of an alkanamine and ammonia with a proton to compare their basicity. Due to the electron releasing nature of alkyl group, it pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid. Moreover, the substituted ammonium ion formed from the amine gets stabilised due to dispersal of the positive charge by the +I effect of the alkyl group. Hence, alkylamines are stronger bases than ammonia. Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups. This trend is followed in the gaseous phase. The order of basicity of amines in the gaseous phase follows the expected order: tertiary amine > secondary amine > primary amine > NH3. The trend is not regular in the aqueous state as evident by their pKb values given in Table 9.3. In the aqueous phase, the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group (+I) but also by solvation with water molecules. The greater the size of the ion, lesser will be the solvation and the less stabilised is the ion. The order of stability of ions are as follows: Decreasing order of extent of H-bonding in water and order of stability of ions by solvation. Chemistry 268 Reprint 2025-26 Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like –CH3 group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than CH3 group, there will be steric hinderance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from –CH3 to –C2H5 results in change of the order of basic strength. Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows: (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (b) Arylamines versus ammonia pKb value of aniline is quite high. Why is it so? It is because in aniline or other arylamines, the -NH2 group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation. If you write different resonating structures of aniline, you will find that aniline is a resonance hybrid of the following five structures. On the other hand, anilinium ion obtained by accepting a proton can have only two resonating structures (kekule). We know that greater the number of resonating structures, greater is the stability. Thus you can infer that aniline (five resonating structures) is more stable than anilinium ion. Hence, the proton acceptability or the basic nature of aniline or other aromatic amines would be less than that of ammonia. In case of substituted aniline, it is observed that electron releasing groups like –OCH3, –CH3 increase basic strength whereas electron withdrawing groups like –NO2, –SO3H, –COOH, –X decrease it. 269 Amines Reprint 2025-26 ExampleExampleExampleExampleExample 9.49.49.49.49.4 Arrange the following in decreasing order of their basic strength: C6H5NH2, C2H5NH2, (C2H5)2NH, NH3 SolutionSolutionSolutionSolutionSolution The decreasing order of basic strength of the above amines and ammonia follows the following order: (C2H5)2NH > C2H5NH2 > NH3 > C6H5NH2 2. Alkylation Amines undergo alkylation on reaction with alkyl halides (refer Unit 6, Class XII). 3. Acylation Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is known as acylation. You can consider this reaction as the replacement of hydrogen atom of –NH2 or >N–H group by the acyl group. The products obtained by acylation reaction are known as amides. The reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right hand side. Amines also react with benzoyl chloride (C6H5COCl). This reaction is known as benzoylation. CH 3 NH 2  C 6 H 5 CO Cl  CH 3 NH CO C 6 H 5  H Cl Methanamine Benzoyl chloride N  Methylbenzamide What do you think is the product of the reaction of amines with carboxylic acids ? They form salts with amines at room temperature. Chemistry 270 Reprint 2025-26 4. Carbylamine reaction Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines. 5. Reaction with nitrous acid Three classes of amines react differently with nitrous acid which is prepared in situ from a mineral acid and sodium nitrite. (a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols. Quantitative evolution of nitrogen is used in estimation of amino acids and proteins. (b) Aromatic amines react with nitrous acid at low temperatures (273-278 K) to form diazonium salts, a very important class of compounds used for synthesis of a variety of aromatic compounds discussed in Section 9.7. Secondary and tertiary amines react with nitrous acid in a different manner. 6. Reaction with arylsulphonyl chloride Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides. (a) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali. (b) In the reaction with secondary amine, N,N-diethyl- benzenesulphonamide is formed. O O S Cl + H N C 2H 5 S N C 2H 5 + HCl O C 2H 5 O C 2H 5 N,N-Diethylbenzenesulphonamide 271 Amines Reprint 2025-26 Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali. (c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines. However, these days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride. 7. Electrophilic substitution You have read earlier that aniline is a resonance hybrid of five structures. Where do you find the maximum electron density in these structures? Ortho- and para-positions to the –NH2 group become centres of high electron density. Thus –NH2 group is ortho and para directing and a powerful activating group. (a) Bromination: Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their very high reactivity. Substitution tends to occur at ortho- and para-positions. If we have to prepare monosubstituted aniline derivative, how can the activating effect of –NH2 group be controlled ? This can be done by protecting the -NH2 group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below: Chemistry 272 Reprint 2025-26 Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of –NHCOCH3 group is less than that of amino group. (b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed. However, by protecting the –NH2 group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product. (c) Sulphonation: Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product. Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. 273 Amines Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 9.4 Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. 9.5 Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl ® (ii) (C2H5)3N + HCl ® 9.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. 9.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. 9.8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. II. DIAZONIUM SALTS  – The diazonium salts have the general formula R N 2 X where R stands – – – for an aryl group and X ion may be Cl Br, HSO,4 BF,4 etc. They are named by suffixing diazonium to the name of the parent hydrocarbon from which they are formed, followed by the name of anion such as  chloride, hydrogensulphate, etc. The N 2 group is called diazonium  – group. For example, C 6 H 5 N 2 Cl is named as benzenediazonium chloride and C6H5N2+HSO4– is known as benzenediazonium hydrogensulphate. Primary aliphatic amines form highly unstable alkyldiazonium salts (refer to Section 9.6). Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273-278 K). The stability of arenediazonium ion is explained on the basis of resonance. 9.79.79.79.79.7 MethodMethodMethodMethodMethod ofofofofof Benzenediazonium chloride is prepared by the reaction of aniline with PreparationPreparationPreparationPreparationPreparation nitrous acid at 273-278K. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid. The ofofofofof DiazoniunDiazoniunDiazoniunDiazoniunDiazoniun conversion of primary aromatic amines into diazonium salts is known SaltsSaltsSaltsSaltsSalts as diazotisation. Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation.  – C 6 H 5 N H 2  NaNO 2  2H Cl 273  278K C 6 H 5 N 2 Cl  Na Cl  2H 2 O Chemistry 274 Reprint 2025-26 9.89.89.89.89.8 PhysicalPhysicalPhysicalPhysicalPhysical Benzenediazonium chloride is a colourless crystalline solid. It is readily soluble in water and is stable in cold but reacts with water when PropertiesPropertiesPropertiesPropertiesProperties warmed. It decomposes easily in the dry state. Benzenediazonium fluoroborate is water insoluble and stable at room temperature. 9.99.99.99.99.9 ChemicalChemicalChemicalChemicalChemical The reactions of diazonium salts can be broadly divided into two ReactionsReactionsReactionsReactionsReactions categories, namely (A) reactions involving displacement of nitrogen and (B) reactions involving retention of diazo group. A. Reactions involving displacement of nitrogen Diazonium group being a very good leaving group, is substituted by other groups such as Cl–, Br –, I – , CN – and OH– which displace nitrogen from the aromatic ring. The nitrogen formed escapes from the reaction mixture as a gas. 1. Replacement by halide or cyanide ion: The Cl –, Br– and CN – nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction. 2 2 2 2 Alternatively, chlorine or bromine can also be introduced in the benzene ring by treating the diazonium salt solution with corresponding halogen acid in the presence of copper powder. This is referred as Gatterman reaction. The yield in Sandmeyer reaction is found to be better than Gattermann reaction. 2. Replacement by iodide ion: Iodine is not easily introduced into the benzene ring directly, but, when the diazonium salt solution is treated with potassium iodide, iodobenzene is formed. 3. Replacement by fluoride ion: When arenediazonium chloride is treated with fluoroboric acid, arene diazonium fluoroborate is precipitated which on heating decomposes to yield aryl fluoride. 4. Replacement by H: Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal, respectively. 275 Amines Reprint 2025-26 5. Replacement by hydroxyl group: If the temperature of the diazonium salt solution is allowed to rise upto 283 K, the salt gets hydrolysed to phenol. 6. Replacement by –NO2 group: When diazonium fluoroborate is heated with aqueous sodium nitrite solution in the presence of copper, the diazonium group is replaced by –NO2 group. B. Reactions involving retention of diazo group coupling reactions The azo products obtained have an extended conjugate system having both the aromatic rings joined through the –N=N– bond. These compounds are often coloured and are used as dyes. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling reaction. Similarly the reaction of diazonium salt with aniline yields p-aminoazobenzene. This is an example of electrophilic substitution reaction. 9.10.10.10.10.10 ImportanceImportanceImportanceImportanceImportance From the above reactions, it is clear that the diazonium salts are very ofofofofof DiazoniumDiazoniumDiazoniumDiazoniumDiazonium good intermediates for the introduction of –F, –Cl, –Br, –I, –CN, –OH, –NO2 groups into the aromatic ring. SaltsSaltsSaltsSaltsSalts ininininin Aryl fluorides and iodides cannot be prepared by direct halogenation. SynthesisSynthesisSynthesisSynthesisSynthesis The cyano group cannot be introduced by nucleophilic substitution of ofofofofof AromaticAromaticAromaticAromaticAromatic chlorine in chlorobenzene but cyanobenzene can be easily obtained from diazonium salt. CompoundsCompoundsCompoundsCompoundsCompounds Thus, the replacement of diazo group by other groups is helpful in Chemistry 276 Reprint 2025-26 preparing those substituted aromatic compounds which cannot be prepared by direct substitution in benzene or substituted benzene. How will you convert 4-nitrotoluene to 2-bromobenzoic acid ? ExampleExampleExampleExampleExample 9.59.59.59.59.5 SolutionSolutionSolutionSolutionSolution IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 9.9 Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5 - tribromobenzene. SummarySummarySummarySummarySummary Amines can be considered as derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. Replacement of one hydrogen atom of ammonia gives rise to structure of the type R-NH2, known as primary amine. Secondary amines are characterised by the structure R2NH or R-NHR¢ and tertiary amines by R3N, RNR¢R¢¢¢¢¢¢¢¢¢¢ or R2NR¢.¢.¢.¢.¢. Secondary and tertiary amines are known as simple amines if the alkyl or aryl groups are the same and mixed amines if the groups are different. Like ammonia, all the three types of amines have one unshared electron pair on nitrogen atom due to which they behave as Lewis bases. Amines are usually formed from nitro compounds, halides, amides, imides, etc. They exhibit hydrogen bonding which influence their physical properties. In alkylamines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. In aromatic amines, electron releasing and withdrawing groups, respectively increase and decrease their basic character. Aniline is a weaker base 277 Amines Reprint 2025-26 than ammonia. Reactions of amines are governed by availability of the unshared pair of electrons on nitrogen. Influence of the number of hydrogen atoms at nitrogen atom on the type of reactions and nature of products is responsible for identification and distinction between primary, secondary and tertiary amines. p-Toluenesulphonyl chloride is used for the identification of primary, secondary and tertiary amines. Presence of amino group in aromatic ring enhances reactivity of the aromatic amines. Reactivity of aromatic amines can be controlled by acylation process, i.e., by treating with acetyl chloride or acetic anhydride. Tertiary amines like trimethylamine are used as insect attractants. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes by reductive removal of the diazo group. Coupling reaction of aryldiazonium salts with phenols or arylamines give rise to the formation of azo dyes. Exercises 9.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2)2NCH3 (vii) m–BrC6H4NH2 9.2 Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline. 9.3 Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v) Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines. 9.4 Arrange the following: (i) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine Chemistry 278 Reprint 2025-26 (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2. (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.

8.6 ISOMERISM isomers and this phenomenon is termed as position isomerism. For example, the The phenomenon of existence of two or more molecular formula C3H8O represents twocompounds possessing the same molecular alcohols: formula but different properties is known as isomerism. Such compounds are called OH as isomers. The following flow chart shows different types of isomerism. CH3CH2CH2OH CH3−CH-CH3 Propan-1-ol Propan-2-ol 8.6.1 Structural Isomerism Compounds having the same molecular (iii) Functional group isomerism: Two or formula but different structures (manners more compounds having the same molecular in which atoms are linked) are classified as formula but different functional groups structural isomers. Some typical examples are called functional isomers and this of different types of structural isomerism are phenomenon is termed as functional group given below: isomerism. For example, the molecular (i) Chain isomerism: When two or more formula C3H6O represents an aldehyde and compounds have similar molecular formula but a ketone: Isomerism Structural isomerism Stereoisomerism Chain Position Functional Metamerism Geometrical Optical isomerism isomerism group isomerism isomerism isomerism Reprint 2025-26 organic chemistry – some basic principles and techniques 271 in understanding the reactivity of organic O H compounds and in planning strategy for their   synthesis. CH3−C-CH3 CH3−CH2—C= O In the following sections, we shall learn Propanone Propanal some of the principles that explain how these (iv) Metamerism: It arises due to different reactions take place. alkyl chains on either side of the functional 8.7.1 Fission of a Covalent Bond group in the molecule. For example, C4H10O represents methoxypropane (CH3OC3H7) and A covalent bond can get cleaved either by : (i) ethoxyethane (C2H5OC2H5). heterolytic cleavage, or by (ii) homolytic cleavage.8.6.2 Stereoisomerism In heterolytic cleavage, the bond breaks The compounds that have the same in such a fashion that the shared pair of constitution and sequence of covalent bonds electrons remains with one of the fragments. but differ in relative positions of their atoms After heterolysis, one atom has a sextet or groups in space are called stereoisomers. electronic structure and a positive charge and This special type of isomerism is called as the other, a valence octet with at least one lone stereoisomerism and can be classified as pair and a negative charge. Thus, heterolytic geometrical and optical isomerism. + cleavage of bromomethane will give C H3 and

8.7 FUNDAMENTAL CONCEPTS IN Br– as shown below. ORGANIC REACTION MECHANISM In an organic reaction, the organic molecule A species having a carbon atom possessing(also referred as a substrate) reacts with an sextext of electrons and a positive charge isappropriate attacking reagent and leads to the formation of one or more intermediate(s) called a carbocation (earlier called carbonium and finally product(s) ion). The H3 ion is known as a methyl cation or methyl carbonium ion. Carbocations areThe general reaction is depicted as follows : classified as primary, secondary or tertiary Attacking depending on whether one, two or three Reagent [Intermediate] Product(s) carbons are directly attached to the positivelyOrganic molecule charged carbon. Some other examples of Byproducts + (Substrate) carbocations are: CH3C H2 (ethyl+ cation, a primary carbocation), (CH3)2C H (isopropyl+ Substrate is that reactant which supplies cation, a secondary carbocation), and (CH3)3C carbon to the new bond and the other reactant (tert-butyl cation, a tertiary carbocation). is called reagent. If both the reactants Carbocations are highly unstable and supply carbon to the new bond then choice reactive species. Alkyl groups directly is arbitrary and in that case the molecule on attached to the positively charged carbon which attention is focused is called substrate. stabilise the carbocations due to inductive In such a reaction a covalent bond and hyperconjugation effects, which you will between two carbon atoms or a carbon and be studying in the sections 8.7.5 and 8.7.9. some other atom is broken and a new bond is The+ observed+ order of carbocation+ stability+ formed. A sequential account of each step, is: C H3 < CH3C H2 < (CH3)2CH < (CH3)3C. These describing details of electron movement, carbocations have trigonal planar shape energetics during bond cleavage and bond with positively charged carbon+ being sp2 formation, and the rates of transformation hybridised. Thus, the shape of C H3 may be of reactants into products (kinetics) is considered as being derived from the overlap referred to as reaction mechanism. The of three equivalent C(sp2) hybridised orbitals knowledge of reaction mechanism helps with 1s orbital of each of the three hydrogen Reprint 2025-26 272 chemistry atoms. Each bond may be represented as curved arrow. Such cleavage results in C(sp2)–H(1s) sigma bond. The remaining the formation of neutral species (atom or carbon orbital is perpendicular to the group) which contains an unpaired electron. molecular plane and contains no electrons. These species are called free radicals. Like [Fig. 8.3(a)]. carbocations and carbanions, free radicals are also very reactive. A homolytic cleavage can be shown as: Alkyl free radical Alkyl radicals are classified as primary, secondary, or tertiary. Alkyl radical stability increases as we proceed from primary to Fig. 8.3 (a) Shape of methyl carbocation tertiary: The heterolytic cleavage can also give a , species in which carbon gets the shared Methyl Ethyl Isopropyl Tert-butyl pair of electrons. For example, when group free free free free Z attached to the carbon leaves without radical radical radical radical Organic reactions, which proceed by homolytic fission are called free radical or electron pair, the methyl anion is homopolar or nonpolar reactions. formed. Such a carbon species carrying a 8.7.2 Substrate and Reagent negative charge on carbon atom is called Ions are generally not formed in the reactions carbanion. Carbon in carbanion is generally of organic compounds. Molecules as such sp3 hybridised and its structure is distorted participate in the reaction. It is convenient to tetrahedron as shown in Fig. 8.3(b). name one reagent as substrate and other as reagent. In general, a molecule whose carbon is involved in new bond formation is called substrate and the other one is called reagent. When carbon-carbon bond is formed, the choice of naming the reactants as substrate and reagent is arbitrary and depends on molecule under observation. Example: Fig. 8.3 (b) Shape of methyl carbanion (i) CH2 = CH2 + Br2 → CH2 Br – CH2Br Substrate Reagent Product Carbanions are also unstable and reactive species. The organic reactions which proceed through heterolytic bond cleavage are called (ii) ionic or heteropolar or just polar reactions. In homolytic cleavage, one of the electrons of the shared pair in a covalent bond goes with each of the bonded atoms. Thus, in homolytic cleavage, the movement of a Nucleophiles and Electrophilessingle electron takes place instead of an electron pair. The single electron movement Reagents attack the reactive site of the is shown by ‘half-headed’ (fish hook: ) substrate. The reactive site may be electron Reprint 2025-26 organic chemistry – some basic principles and techniques 273 deficient portion of the molecule (a positive Problem 8.11reactive site) e.g., an atom with incomplete electron shell or the positive end of the dipole Using curved-arrow notation, show the in the molecule. If the attacking species is formation of reactive intermediates when electron rich, it attacks these sites. If attacking the following covalent bonds undergo heterolytic cleavage.species is electron deficient, the reactive site for it is that part of the substrate molecule (a) CH3–SCH3, (b) CH3–CN, (c) CH3–Cu which can supply electrons, e.g., π electrons Solution in a double bond. A reagent that brings an electron pair to the reactive site is called a nucleophile (Nu:) i.e., nucleus seeking and the reaction is then called nucleophilic. A reagent that takes away an electron pair from reactive site is called electrophile (E+) i.e., electron seeking and the reaction is called electrophilic. Problem 8.12 During a polar organic reaction, a Giving justification, categorise the nucleophile attacks an electrophilic centre following molecules/ions as nucleophile of the substrate which is that specific atom or electrophile: or part of the substrate which is electron deficient. Similarly, the electrophiles attack at nucleophilic centre, which is the electron rich centre of the substrate. Thus, the electrophiles receive electron pair from the Solution substrate when the two undergo bonding Nucleophiles: HS–,C2H5O–,(CH3)3N:H2N:– interaction. A curved-arrow notation is used These species have unshared pair of to show the movement of an electron pair electrons, which can be donated and from the nucleophile to the electrophile. Some shared with an electrophile. examples of nucleophiles are the negatively + + Electrophiles: BF3,C1+ H3–C =O,N O2.charged ions with lone pair of electrons such Reactive sites have only six valence as hydroxide (HO– ), cyanide (NC–) ions and electrons; can accept electron pair from carbanions (R3C:–). Neutral molecules such a nucleophile. as etc., can also act as Problem 8.13nucleophiles due to the presence of lone Identify electrophilic centre in thepair of electrons. Examples of electrophiles + following: CH3CH=O, CH3CN, CH3I.include carbocations (C H3) and neutral molecules having functional groups like Solution carbonyl group (>C=O) or alkyl halides Among CH 3HC*=O, H 3CC*≡N, and (R3C-X, where X is a halogen atom). The H3C*–I, the starred carbon atoms arecarbon atom in carbocations has sextet electrophilic centers as they will haveconfiguration; hence, it is electron deficient partial positive charge due to polarity ofand can receive a pair of electrons from the the bond.nucleophiles. In neutral molecules such as alkyl halides, due to the polarity of the C-X 8.7.3 Electron Movement in Organicbond a partial positive charge is generated Reactionson the carbon atom and hence the carbon atom becomes an electrophilic centre at which The movement of electrons in organic a nucleophile can attack. reactions can be shown by curved-arrow Reprint 2025-26 274 chemistry notation. It shows how changes in bonding 8.7.5 Inductive Effect occur due to electronic redistribution during When a covalent bond is formed between the reaction. To show the change in position atoms of different electronegativity, the of a pair of electrons, curved arrow starts electron density is more towards the more from the point from where an electron pair is electronegative atom of the bond. Such a shift shifted and it ends at a location to which the of electron density results in a polar covalent pair of electron may move. bond. Bond polarity leads to various electronic Presentation of shifting of electron pair is effects in organic compounds. given below : Let us consider cholorethane (CH3CH2Cl) in which the C–Cl bond is a polar covalent(i) from π bond to bond. It is polarised in such a way that the adjacent bond position carbon-1 gains some positive charge (δ+) (ii) from π bond to and the chlorine some negative charge (δ–). adjacent atom The fractional electronic charges on the two (iii) from atom to adjacent atoms in a polar covalent bond are denoted bond position by symbol δ (delta) and the shift of electron density is shown by an arrow that points from Movement of single electron is indicated by δ+ to δ– end of the polar bond.a single barbed ‘fish hooks’ (i.e. half headed curved arrow). For example, in transfer of δδ+ δ+ δ− hydroxide ion giving ethanol and in the CH3→CH2→Cl dissociation of chloromethane, the movement 2 1 In turn carbon-1, which has developedof electron using curved arrows can be partial positive charge (δ+) draws somedepicted as follows: electron density towards it from the adjacent C-C bond. Consequently, some positive charge (δδ+) develops on carbon-2 also, where δδ+ symbolises relatively smaller positive charge as compared to that on carbon – 1. In other words, the polar C – Cl bond induces polarity in the adjacent bonds. Such polarisation of8.7.4 Electron Displacement Effects in σ-bond caused by the polarisation of adjacent Covalent Bonds σ-bond is referred to as the inductive effect. The electron displacement in an organic This effect is passed on to the subsequent molecule may take place either in the ground bonds also but the effect decreases rapidly state under the influence of an atom or a as the number of intervening bonds increases substituent group or in the presence of an and becomes vanishingly small after three appropriate attacking reagent. The electron bonds. The inductive effect is related to the displacements due to the influence of ability of substituent(s) to either withdraw or an atom or a substituent group present in donate electron density to the attached carbon the molecule cause permanent polarlisation atom. Based on this ability, the substitutents of the bond. Inductive effect and resonance can be classified as electron-withdrawing or effects are examples of this type of electron electron donating groups relative to hydrogen. displacements. Temporary electron Halogens and many other groups such as displacement effects are seen in a molecule nitro (- NO2), cyano (- CN), carboxy (- COOH), when a reagent approaches to attack it. ester (COOR), aryloxy (-OAr, e.g. – OC6H5), etc. This type of electron displacement is called are electron-withdrawing groups. On the other electromeric effect or polarisability effect. hand, the alkyl groups like methyl (–CH3) and In the following sections we will learn about ethyl (–CH2–CH3) are usually considered as these types of electronic displacements. electron donating groups. ­ Reprint 2025-26 organic chemistry – some basic principles and techniques 275 benzene cannot be adequately represented Problem 8.14 by any of these structures, rather it is Which bond is more polar in the following a hybrid of the two structures (I and II) pairs of molecules: (a) H3C-H, H3C-Br called resonance structures. The resonance (b) H3C-NH2, H3C-OH (c) H3C-OH, structures (canonical structures or H3C-SH contributing structures) are hypothetical and individually do not represent any Solution real molecule. They contribute to the actual (a) C–Br, since Br is more electronegative structure in proportion to their stability. than H, (b) C–O, (c) C–O Another example of resonance is provided Problem 8.15 by nitromethane (CH3NO2) which can be represented by two Lewis structures, (I and In which C–C bond of CH3CH2CH2Br, the II). There are two types of N-O bonds in these inductive effect is expected to be the least? structures. Solution Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence, the effect is least in the bond between carbon-3 and hydrogen. However, it is known that the two N–O bonds of nitromethane are of the same8.7.6 Resonance Structure length (intermediate between a N–O single There are many organic molecules whose bond and a N=O double bond). The actual behaviour cannot be explained by a single Lewis structure of nitromethane is therefore structure. An example is that of a resonance hybrid of the two canonical benzene. Its cyclic structure forms I and II. containing alternating C–C single The energy of actual structure of the molecule and C=C double bonds shown (the resonance hybrid) is lower than that of is inadequate for explaining its Benzene any of the canonical structures. The difference characteristic properties. in energy between the actual structure and the As per the above representation, benzene lowest energy resonance structure is called the should exhibit two different bond lengths, resonance stabilisation energy or simply the resonance energy. The more the numberdue to C–C single and C=C double bonds. of important contributing structures, theHowever, as determined experimentally more is the resonance energy. Resonance isbenzene has a uniform C–C bond distances particularly important when the contributingof 139 pm, a value intermediate between the structures are equivalent in energy. C–C single(154 pm) and C=C double (134 The following rules are applied while writingpm) bonds. Thus, the structure of benzene resonance structures:cannot be represented adequately by the The resonance structures have (i) the sameabove structure. Further, benzene can be positions of nuclei and (ii) the same number of represented equally well by the energetically unpaired electrons. Among the resonance identical structures I and II. structures, the one which has more number of covalent bonds, all the atoms with octet of electrons (except hydrogen which has a duplet), less separation of opposite charges, (a negative charge if any on more electronegative atom, a positive charge if any on more Therefore, according to the resonance electropositive atom) and more dispersal of theory (Unit 4) the actual structure of charge, is more stable than others. Reprint 2025-26 276 chemistry Problem 8.16 Solution Write resonance structures of CH3COO– The two structures are less important and show the movement of electrons by contributors as they involve charge curved arrows. separation. Additionally, structure I contains a carbon atom with an Solution incomplete octet. First, write the structure and put unshared pairs of valence electrons on 8.7.7 Resonance Effect appropriate atoms. Then draw the arrows The resonance effect is defined as ‘the polarity one at a time moving the electrons to get produced in the molecule by the interaction the other structures. of two π-bonds or between a π-bond and lone pair of electrons present on an adjacent atom’. The effect is transmitted through the chain. There are two types of resonance or mesomeric effect designated as R or M effect. Problem 8.17 (i) Positive Resonance Effect (+R effect) Write resonance structures of In this effect, the transfer of electrons is away CH2=CH–CHO. Indicate relative stability from an atom or substituent group attached of the contributing structures. to the conjugated system. This electron displacement makes certain positions in the Solution molecule of high electron densities. This effect in aniline is shown as : Stability: I > II > III (ii) Negative Resonance Effect (- R effect) [I: Most stable, more number of covalent This effect is observed when the transfer of bonds, each carbon and oxygen atom has electrons is towards the atom or substituent an octet and no separation of opposite group attached to the conjugated system. charge II: negative charge on more For example in nitrobenzene this electron electronegative atom and positive charge displacement can be depicted as : on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable]. Problem 8.18 Explain why the following two structures, I and II cannot be the major contributors The atoms or substituent groups, which to the real structure of CH3COOCH3. represent +R or –R electron displacement effects are as follows : +R effect: – halogen, –OH, –OR, –OCOR, –NH2, –NHR, –NR2, –NHCOR, – R effect: – COOH, –CHO, >C=O, – CN, –NO2 Reprint 2025-26 organic chemistry – some basic principles and techniques 277 The presence of alternate single and system or to an atom with an unshared double bonds in an open chain or cyclic system p orbital. The σ electrons of C—H bond of the is termed as a conjugated system. These alkyl group enter into partial conjugation with systems often show abnormal behaviour. the attached unsaturated system or with the The examples are 1,3- butadiene, aniline unshared p orbital. Hyperconjugation is a and nitrobenzene etc. In such systems, the permanent effect. π-electrons are delocalised and the system To understand hyperconjugation effect, let develops polarity. + us take an example of CH3 C H2 (ethyl cation) in 8.7.8 Electromeric Effect (E effect) which the positively charged carbon atom has an empty p orbital. One of the C-H bonds ofIt is a temporary effect. The organic compounds the methyl group can align in the plane of thishaving a multiple bond (a double or triple empty p orbital and the electrons constitutingbond) show this effect in the presence of the C-H bond in plane with this p orbital canan attacking reagent only. It is defined as then be delocalised into the empty p orbitalthe complete transfer of a shared pair of as depicted in Fig. 8.4 (a).π-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. The effect is annulled as soon as the attacking reagent is removed from the domain of the reaction. It is represented by E and the shifting of the electrons is shown by a curved arrow ( ). There are two distinct types of electromeric effect. (i) Positive Eelctromeric Effect (+E effect) In this effect the π−electrons of the multiple bond are transferred to that atom to which the reagent gets attached. For example: Fig. 8.4(a) Orbital diagram showing hyperconjugation in ethyl cation This type of overlap stabilises the carbocation because electron density from (ii) Negative Electromeric Effect (–E effect) In the adjacent σ bond helps in dispersing the this effect the π - electrons of the multiple positive charge. bond are transferred to that atom to which the attacking reagent does not get attached. For example: When inductive and electromeric effects operate in opposite directions, the electomeric effect predominates. In general, greater the number of alkyl 8.7.9 Hyperconjugation groups attached to a positively charged carbon Hyperconjugation is a general stabilising atom, the greater is the hyperconjugation interaction. It involves delocalisation of interaction and stabilisation of the cation. σ electrons of C—H bond of an alkyl group Thus, we have the following relative stability directly attached to an atom of unsaturated of carbocations : Reprint 2025-26 278 chemistry Problem 8.19 + Explain why (CH3)3C is more stable + + than CH3C H2 and C H3 is the least stable cation. Hyperconjugation is also possible in Solution alkenes and alkylarenes. + Hyperconjugation interaction in (CH3)3C is+ + D e l o c a l i s a t i o n o f e l e c t r o n s b y greater than in CH3C H2 as+ the (CH3)3Chyperconjugation in the case of alkene can has nine C-H bonds. In C H3, vacant pbe depicted as in Fig. 8.4(b). orbital is perpendicular to the plane in which C-H bonds lie; hence cannot + overlap with it. Thus, C H3 lacks hyperconjugative stability. 8.7.10 Types of Organic Reactions and Mechanisms Organic reactions can be classified into the Fig. 8.4(b) Orbital diagram showing following categories: hyperconjugation in propene (i) Substitution reactions (ii) Addition reactions There are various ways of looking at the (iii) Elimination reactions hyperconjugative effect. One of the way is to (iv) Rearrangement reactions regard C—H bond as possessing partial ionic You will be studying these reactions incharacter due to resonance. Unit 9 and later in class XII. 8.8 Methods of Purification of Organic Compounds Once an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it. The common techniques used for purification are as follows : (i) Sublimation (ii) Crystallisation (iii) Distillation (iv) Differential extraction and (v) Chromatography Finally, the purity of a compound is ascertained by determining its melting or boiling point. Most of the pure compounds have sharp melting points and boiling points. The hyperconjugation may also be New methods of checking the purity of an regarded as no bond resonance. organic compound are based on different Reprint 2025-26 organic chemistry – some basic principles and techniques 279 types of chromatographic and spectroscopic carefully. On boiling, the vapours of lower techniques. boiling component are formed first. The vapours are condensed by using a condenser8.8.1 Sublimation and the liquid is collected in a receiver. You have learnt earlier that on heating, some The vapours of higher boiling component solid substances change from solid to vapour form later and the liquid can be collected state without passing through liquid state. separately. The purification technique based on the above principle is known as sublimation and is used to separate sublimable compounds from non- sublimable impurities. 8.8.2 Crystallisation This is one of the most commonly used techniques for the purification of solid organic compounds. It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. On cooling the solution, pure compound crystallises out and is removed by filtration. The filtrate (mother liquor) contains impurities and small quantity of the compound. If the compound is highly soluble in one solvent and very little soluble in another solvent, crystallisation can be Fig.8.5 Simple distillation. The vapours of a substance formed are condensed andsatisfactorily carried out in a mixture of these the liquid is collected in conical flask. solvents. Impurities, which impart colour to the solution are removed by adsorbing over Fractional Distillation: If the differenceactivated charcoal. Repeated crystallisation in boiling points of two liquids is not much,becomes necessary for the purification of compounds containing impurities of simple distillation cannot be used to separate comparable solubilities. them. The vapours of such liquids are formed within the same temperature range and are 8.8.3 Distillation condensed simultaneously. The technique of This important method is used to separate fractional distillation is used in such cases. In (i) volatile liquids from nonvolatile impurities this technique, vapours of a liquid mixture are and (ii) the liquids having sufficient difference passed through a fractionating column before in their boiling points. Liquids having condensation. The fractionating column is different boiling points vaporise at different fitted over the mouth of the round bottom temperatures. The vapours are cooled and flask (Fig.8.6, page 280). the liquids so formed are collected separately. Vapours of the liquid with higher boiling Chloroform (b.p 334 K) and aniline (b.p. 457 point condense before the vapours of the K) are easily separated by the technique of liquid with lower boiling point. The vapours distillation (Fig 8.5). The liquid mixture is rising up in the fractionating column become taken in a round bottom flask and heated richer in more volatile component. By the Reprint 2025-26 280 chemistry unit in the fractionating column is called a theoretical plate. Commercially, columns with hundreds of plates are available. One of the technological applications of fractional distillation is to separate different fractions of crude oil in petroleum industry. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water pump or vacuum pump (Fig.8.8). Glycerol can be separated from Fig.8.6 Fractional distillation. The vapours of lower boiling spent-lye in soap industry by fraction reach the top of the column first followed by using this technique. vapours of higher boiling fractions. time the vapours reach to the top of the fractionating column, these are rich in the more volatile component. Fractionating columns are available in various sizes and designs as shown in Fig.8.7. A fractionating column provides many surfaces for heat exchange between the ascending vapours and the descending condensed liquid. Some of the condensing liquid in the fractionating column obtains heat from the ascending vapours and revaporises. The vapours thus become richer in low boiling component. The vapours of low boiling component ascend to the top of the column. On reaching the top, the vapours become pure in low boiling component and pass through the condenser and the pure liquid is collected in a receiver. After a series of successive distillations, the remaining liquid in the distillation flask gets enriched in high boiling component. Each Fig.8.7 Different types of fractionating columns.successive condensation and vaporisation Reprint 2025-26 organic chemistry – some basic principles and techniques 281 Fig.8.8 Distillation under reduced pressure. A liquid boils at a temperature below its vapour pressure by reducing the pressure. Steam Distillation: This technique is 8.8.4 Differential Extraction applied to separate substances which are When an organic compound is present in an steam volatile and are immiscible with water. aqueous medium, it is separated by shaking In steam distillation, steam from a steam it with an organic solvent in which it is more generator is passed through a heated flask soluble than in water. The organic solvent and containing the liquid to be distilled. The the aqueous solution should be immiscible mixture of steam and the volatile organic with each other so that they form two distinct compound is condensed and collected. The layers which can be separated by separatory compound is later separated from water using funnel. The organic solvent is later removed a separating funnel. In steam distillation, by distillation or by evaporation to get back the liquid boils when the sum of vapour the compound. Differential extraction is pressures due to the organic liquid (p1) carried out in a separatory funnel as shown in and that due to water (p2) becomes equal to Fig. 8.10 (Page 282). If the organic compound the atmospheric pressure (p), i.e. p =p1+ p2. is less soluble in the organic solvent, a very Since p1 is lower than p, the organic liquid large quantity of solvent would be required vaporises at lower temperature than its to extract even a very small quantity of the boiling point. compound. The technique of continuous extraction is employed in such cases. In this Thus, if one of the substances in the technique same solvent is repeatedly used formixture is water and the other, a water extraction of the compound.insoluble substance, then the mixture will boil close to but below, 373K. A mixture of water 8.8.5 Chromatography and the substance is obtained which can Chromatography is an important technique be separated by using a separating funnel. extensively used to separate mixtures into Aniline is separated by this technique from their components, purify compounds and aniline – water mixture (Fig.8.9, Page 282). also to test the purity of compounds. The Reprint 2025-26 282 chemistry Fig.8.9 Steam distillation. Steam volatile component volatilizes, the vapours condense in the condenser and the liquid collects in conical flask. name chromatography is based on the Greek word chroma, for colour since the method was first used for the separation of coloured substances found in plants. In this technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the mixture get gradually separated from one another. The moving phase is called the mobile phase. Based on the principle involved, chromatography is classified into different categories. Two of these are: (a) Adsorption chromatography, and Fig.8.10 Differential extraction. Extraction of com- (b) Partition chromatography. pound takes place based on difference in solubility a) Adsorption Chromatography: Adsor- ption chromatography is based on the fact distances over the stationary phase. Followingthat different compounds are adsorbed on are two main types of chromatographican adsorbent to different degrees. Commonly techniques based on the principle of differentialused adsorbents are silica gel and alumina. adsorption.When a mobile phase is allowed to move over a stationary phase (adsorbent), the (a) Column chromatography, and components of the mixture move by varying (b) Thin layer chromatography. Reprint 2025-26 organic chemistry – some basic principles and techniques 283 Column Chromatography: Column The glass plate is then placed in a closed jar chromatography involves separation of containing the eluant (Fig. 8.12a). As the a mixture over a column of adsorbent eluant rises up the plate, the components of (stationary phase) packed in a glass tube. the mixture move up along with the eluant to The column is fitted with a stopcock at its different distances depending on their degree lower end (Fig. 8.11). The mixture adsorbed of adsorption and separation takes place. on adsorbent is placed on the top of the The relative adsorption of each component adsorbent column packed in a glass tube. of the mixture is expressed in terms of its An appropriate eluant which is a liquid or a retardation factor i.e. Rf value (Fig.8.12 b). mixture of liquids is allowed to flow down the Rf = Distance moved by the substance from base line (x) column slowly. Depending upon the degree to Distance moved by the solvent from base line (y) which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column (Fig.8.11). Fig.8.12 (a) Thin layer chromatography. Chromatogram being developed. Fig.8.11 Column chromatography. Different stages of separation of components Fig.8.12 (b) Developed chromatogram. of a mixture. The spots of coloured compounds are Thin Layer Chromatography: Thin layer visible on TLC plate due to their original colour. chromatography (TLC) is another type of The spots of colourless compounds, which are adsorption chromatography, which involves invisible to the eye but fluoresce in ultraviolet separation of substances of a mixture over light, can be detected by putting the plate under a thin layer of an adsorbent coated on glass ultraviolet light. Another detection technique plate. A thin layer (about 0.2mm thick) of is to place the plate in a covered jar containing an adsorbent (silica gel or alumina) is spread a few crystals of iodine. Spots of compounds, over a glass plate of suitable size. The plate which adsorb iodine, will show up as brown is known as thin layer chromatography plate spots. Sometimes an appropriate reagent or chromaplate. The solution of the mixture may also be sprayed on the plate. For example, to be separated is applied as a small spot amino acids may be detected by spraying the about 2 cm above one end of the TLC plate. plate with ninhydrin solution (Fig.8.12b). Reprint 2025-26 284 chemistry Partition Chromatography: Partition spot on the chromatogram. The spots of the chromatography is based on continuous separated colourless compounds may be differential partitioning of components of observed either under ultraviolet light or by a mixture between stationary and mobile the use of an appropriate spray reagent as phases. Paper chromatography is a type discussed under thin layer chromatography. of partition chromatography. In paper 8.9 Qualitative Analysis of Organic chromatography, a special quality paper Compounds known as chromatography paper is used. The elements present in organic compoundsChromatography paper contains water trapped are carbon and hydrogen. In addition toin it, which acts as the stationary phase. these, they may also contain oxygen, nitrogen, A strip of chromatography paper spotted sulphur, halogens and phosphorus.at the base with the solution of the mixture is suspended in a suitable solvent or a mixture 8.9.1 Detection of Carbon and Hydrogen of solvents (Fig. 8.13). This solvent acts as the Carbon and hydrogen are detected by heating mobile phase. The solvent rises up the paper the compound with copper(II) oxide. Carbon by capillary action and flows over the spot. The present in the compound is oxidised to paper selectively retains different components carbon dioxide (tested with lime-water, which according to their differing partition in the develops turbidity) and hydrogen to water two phases. The paper strip so developed is (tested with anhydrous copper sulphate, known as a chromatogram. The spots of the which turns blue). separated coloured compounds are visible at C + 2CuO 2Cu + CO2different heights from the position of initial 2H + CuO Cu + H2O CO2 + Ca(OH)2 CaCO3↓ + H2O 5H2O + CuSO4 CuSO4.5H2O White Blue 8.9.2 Detection of Other Elements Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place: Na + C + N NaCN 2Na + S Na2S Na + X Na X (X = Cl, Br or I) C, N, S and X come from organic compound. Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract. (A) Test for Nitrogen Fig.8.13 Paper chromatography. Chromatography paper in two different The sodium fusion extract is boiled with shapes. iron(II) sulphate and then acidified with Reprint 2025-26 organic chemistry – some basic principles and techniques 285 concentrated sulphuric acid. The formation bromine and a yellow precipitate, insoluble of Prussian blue colour confirms the presence in ammonium hydroxide shows the presence of nitrogen. Sodium cyanide first reacts of iodine. with iron(II) sulphate and forms sodium X– + Ag+ → AgX hexacyanidoferrate(II). On heating with X represents a halogen – Cl, Br or I. concentrated sulphuric acid some iron(II) If nitrogen or sulphur is also present in the ions are oxidised to iron(III) ions which compound, the sodium fusion extract is react with sodium hexacyanidoferrate(II) first boiled with concentrated nitric acid to to produce iron(III) hexacyanidoferrate(II) decompose cyanide or sulphide of sodium (ferriferrocyanide) which is Prussian blue in formed during Lassaigne’s test. These ions colour. would otherwise interfere with silver nitrate 6CN– + Fe2+ → [Fe(CN)6]4– test for halogens. 3[Fe(CN)6]4– + 4Fe3+ Fe4[Fe(CN)6]3.xH2O (D) Test for Phosphorus Prussian blue The compound is heated with an oxidising (B) Test for Sulphur agent (sodium peroxide). The phosphorus (a) The sodium fusion extract is acidified present in the compound is oxidised to with acetic acid and lead acetate is added phosphate. The solution is boiled with nitric to it. A black precipitate of lead sulphide acid and then treated with ammonium indicates the presence of sulphur. molybdate. A yellow colouration or precipitate indicates the presence of phosphorus. S2– + Pb2+ → PbS Black Na3PO4 + 3HNO3 → H3PO4+3NaNO3 (b) On treating sodium fusion extract with H3PO4 + 12(NH4)2MoO4 + 21HNO3 → sodium nitroprusside, appearance of Ammonium a violet colour further indicates the molybdate presence of sulphur. (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O AmmoniumS2– + [Fe(CN)5NO]2– → [Fe(CN)5NOS]4– phosphomolybdate Violet In case, nitrogen and sulphur both are 8.10 Quantitative Analysis present in an organic compound, sodium Quantitative analysis of compounds is very thiocyanate is formed. It gives blood red colour important in organic chemistry. It helps and no Prussian blue since there are no free chemists in the determination of mass per cyanide ions. cent of elements present in a compound. You Na + C + N + S → NaSCN have learnt in Unit-1 that mass per cent of Fe3+ +SCN– → [Fe(SCN)]2+ elements is required for the determination of Blood red emperical and molecular formula. If sodium fusion is carried out with excess The percentage composition of elements of sodium, the thiocyanate decomposes to present in an organic compound is determined yield cyanide and sulphide. These ions give by the following methods: their usual tests. 8.10.1 Carbon and Hydrogen NaSCN + 2Na → NaCN+Na2S (C) Test for Halogens Both carbon and hydrogen are estimated in one experiment. A known mass of an organicThe sodium fusion extract is acidified with compound is burnt in the presence of excessnitric acid and then treated with silver nitrate. of oxygen and copper(II) oxide. Carbon andA white precipitate, soluble in ammonium hydrogen in the compound are oxidised tohydroxide shows the presence of chlorine, carbon dioxide and water respectively.a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O Reprint 2025-26 286 chemistry Fig.8.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of substance are absorbed in anhydrous calcium chloride and potassium hydroxide solutions respectively contained in U tubes. The mass of water produced is determined 8.10.2 Nitrogen by passing the mixture through a weighed There are two methods for estimation of U-tube containing anhydrous calcium chloride. nitrogen: (i) Dumas method and (ii) Kjeldahl’s Carbon dioxide is absorbed in another U-tube method. containing concentrated solution of potassium (i) Dumas method: The nitrogen containing hydroxide. These tubes are connected in series organic compound, when heated with copper (Fig. 8.14). The increase in masses of calcium oxide in an atmosphere of carbon dioxide, chloride and potassium hydroxide gives the yields free nitrogen in addition to carbon amounts of water and carbon dioxide from dioxide and water. which the percentages of carbon and hydrogen CxHyNz + (2x + y/2) CuO →are calculated. Let the mass of organic compound be x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu m g, mass of water and carbon dioxide Traces of nitrogen oxides formed, if produced be m1 and m2 g respectively; any, are reduced to nitrogen by passing the 12 × m2 × 100 gaseous mixture over a heated copper gauze.Percentage of carbon= 44 × m The mixture of gases so produced is collected over an aqueous solution of potassium 2 × m1 × 100Percentage of hydrogen = hydroxide which absorbs carbon dioxide. 18 × m Nitrogen is collected in the upper part of the Problem 8.20 graduated tube (Fig.8.15). On complete combustion, 0.246 g of an Let the mass of organic compound = m g organic compound gave 0.198g of carbon Volume of nitrogen collected = V1 mL dioxide and 0.1014g of water. Determine Room temperature = T1K the percentage composition of carbon and hydrogen in the compound. PV1 1 × 273 Volumeof nitrogen atSTP = Solution 760 × T1 12 × 0.198 × 100 (Letit be V mL) Percentageof carbon = 44 × 0.246 Where p1 and V1 are the pressure and = 21.95% volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas 2 × 0.1014 × 100 is collected. The value of p1 is obtained by Percentage of hydrogen = 18 × 0.246 the relation; = 4.58% p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g. Reprint 2025-26 organic chemistry – some basic principles and techniques 287 Fig. 8.15 Dumas method. The organic compound yields nitrogen gas on heating it with copper(II) oxide in the presence of CO2 gas. The mixture of gases is collected over potassium hydroxide solution in which CO2 is absorbed and volume of nitrogen gas is determined. 28 × V V mL N 2 atSTP weighs = g 28 × 41.9 22400 41.9mLof nitrogen weighs = g 22400 28 × V × 100 Percentage of nitrogen = 28 × 41.9 × 100 22400 × m Percentageof nitrogen = 22400 × 0.3 Problem 8.21 = 17.46% In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K (ii) Kjeldahl’s method: The compound temperature and 715mm pressure. containing nitrogen is heated with concentrated Calculate the percentage composition sulphuric acid. Nitrogen in the compound of nitrogen in the compound. (Aqueous gets converted to ammonium sulphate tension at 300K=15 mm) (Fig. 8.16). The resulting acid mixture is then Solution heated with excess of sodium hydroxide. Volume of nitrogen collected at 300K and The liberated ammonia gas is absorbed in 715mm pressure is 50 mL an excess of standard solution of sulphuric Actual pressure = 715-15 =700 mm acid. The amount of ammonia produced is 273 × 700 × 50 determined by estimating the amount of Volumeof nitrogen atSTP sulphuric acid consumed in the reaction. It 300 × 760 is done by estimating unreacted sulphuric 41.9mL acid left after the absorption of ammonia by 22,400 mL of N2 at STP weighs = 28 g titrating it with standard alkali solution. The difference between the initial amount of acid Reprint 2025-26 288 chemistry Fig.8.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO4 to get ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed in known volume of standard acid. taken and that left after the reaction gives the 14 × M × 2(V − V1 / 2) 100amount of acid reacted with ammonia. Percentageof N = × 1000 m Organic compound + H2SO4 → (NH4)2SO4 1.4 × M × 2(V − V / 2) = Na2SO4 + 2NH3 + 2H2O m 2NH3 + H2SO4 → (NH4)2SO4 Kjeldahl method is not applicable to Let the mass of organic compound taken = m g compounds containing nitrogen in nitro and Volume of H2SO4 of molarity, M, azo groups and nitrogen present in the ring taken = V mL (e.g. pyridine) as nitrogen of these compounds Volume of NaOH of molarity, M, used for does not change to ammonium sulphate titration of excess of H2SO4 = V1 mL under these conditions. V1mL of NaOH of molarity M Problem 8.22 = V1 /2 mL of H2SO4 of molarity M During estimation of nitrogen presentVolume of H2SO4 of molarity M unused in an organic compound by Kjeldahl’s= (V - V1/2) mL method, the ammonia evolved from (V- V1/2) mL of H2SO4 of molarity M 0.5 g of the compound in Kjeldahl’s = 2(V-V1/2) mL of NH3 solution of estimation of nitrogen, neutralized 10 mL molarity M. of 1 M H2SO4. Find out the percentage of 1000 mL of 1 M NH3 solution contains nitrogen in the compound. 17g NH3 or 14 g of N Solution 2(V-V1/2) mL of NH3 solution of molarity M 1 M of 10 mL H2SO4=1M of 20 mL NH3contains: 1000 mL of 1M ammonia contains 14 g 14 × M × 2(V − V1 / 2) nitrogen g N 1000 20 mL of 1M ammonia contains Reprint 2025-26 organic chemistry – some basic principles and techniques 289 14 × 20 Percentage of halogen g nitrogen 1000 atomicmassof X × m1g = 14×20×100 molecular massof AgX Percentageof nitrogen = =56.0% 1000×0.5 Problem 8.23 8.10.3 Halogens In Carius method of estimation of Carius method: A known mass of an organic halogen, 0.15 g of an organic compound compound is heated with fuming nitric acid in gave 0.12 g of AgBr. Find out the the presence of silver nitrate contained in a hard percentage of bromine in the compound. glass tube known as Carius tube, (Fig.8.17) Solution Molar mass of AgBr = 108 + 80 = 188 g mol-1 188 g AgBr contains 80 g bromine 80 × 0.12 0.12 g AgBr contains g bromine 188 80 × 0.12 × 100 Percentage of bromine = 188×0.15 = 34.04% 8.10.4 Sulphur A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be Fig. 8.17 Carius method. Halogen containing calculated from the mass of barium sulphate. organic compound is heated with fuming Let the mass of organic nitric acid in the presence of silver nitrate. compound taken = m g and the mass of barium in a furnace. Carbon and hydrogen present in sulphate formed = m1g the compound are oxidised to carbon dioxide 1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur and water. The halogen present forms the 32 × m1 corresponding silver halide (AgX). It is filtered, m1 g BaSO4 contains g sulphur 233 × washed, dried and weighed. 32 × m1 × 100Let the mass of organic Percentageof sulphur = compound taken = m g 233 × m Mass of AgX formed = m1 g 1 mol of AgX contains 1 mol of X Problem 8.24Mass of halogen in m1g of AgX atomicmassof X × m1g In sulphur estimation, 0.157 g of an = organic compound gave 0.4813 g of molecular massof AgX Reprint 2025-26 290 chemistry percentage composition (100) and the sum of barium sulphate. What is the percentage the percentages of all other elements. However, of sulphur in the compound? oxygen can also be estimated directly as follows: Solution A definite mass of an organic compound is Molecular mass of BaSO4 = 137+32+64 decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing = 233 g oxygen is passed over red-hot coke when all 233 g BaSO4 contains 32 g sulphur the oxygen is converted to carbon monoxide. 32 × 0.4813 This mixture is passed through warm iodine 0.4813 g BaSO4 contains g pentoxide (I2O5) when carbon monoxide is g sulphur 233 oxidised to carbon dioxide producing iodine. 32 × 0.4813 × 100 Compound heat → O2 + other gaseous Percentageof sulphur = products 233 × 0.157 1373 K 2C + O2 → 2CO]× 5 (A) = 42.10% I2O5 + 5CO → I2 + 5CO2]× 2 (B) 8.10.5 Phosphorus On making the amount of CO produced in equation (A) equal to the amount of CO usedA known mass of an organic compound is in equation (B) by multiplying the equationsheated with fuming nitric acid whereupon (A) and (B) by 5 and 2 respectively; we findphosphorus present in the compound is that each mole of oxygen liberated fromoxidised to phosphoric acid. It is precipitated the compound will produce two moles ofas ammonium phosphomolybdate, (NH4)3 PO4.12MoO3, by adding ammonia and carbondioxide. ammonium molybdate. Alternatively, Thus 88 g carbon dioxide is obtained if 32 g phosphoric acid may be precipitated as oxygen is liberated. MgNH4PO4 by adding magnesia mixture which Let the mass of organic compound taken be m g on ignition yields Mg2P2O7. Mass of carbon dioxide produced be m1 g Let the mass of organic compound taken ∴ m1 g carbon dioxide is obtained from = m g and mass of ammonium phospho 32 × m1 g O2molydate = m1g 88 Molar mass of (NH4)3PO4.12MoO3 = 1877g 32 × m1 × 100 ∴Percentage of oxygen = % 31 × m1 × 100 88 × mPercentage of phosphorus = % 1877 × m The percentage of oxygen can be derived from the amount of iodine produced also. If phosphorus is estimated as Mg2P2O7, Presently, the estimation of elements in 62 × m1 × 100 an organic compound is carried out by usingPercentage of phosphorus = × 222 microquantities of substances and automatic experimental techniques. The elements,where, 222 u is the molar mass of Mg2P2O7, carbon, hydrogen and nitrogen present in am, the mass of organic compound taken, compound are determined by an apparatusm1, the mass of Mg2P2O7 formed and 62, the known as CHN elemental analyser. Themass of two phosphorus atoms present in the analyser requires only a very small amountcompound Mg2P2O7. of the substance (1-3 mg) and displays the 8.10.6 Oxygen values on a screen within a short time. A The percentage of oxygen in an organic compound detailed discussion of such methods is beyond is usually found by difference between the total the scope of this book. Reprint 2025-26 organic chemistry – some basic principles and techniques 291 Summary In this unit, we have learnt some basic concepts in structure and reactivity of organic compounds, which are formed due to covalent bonding. The nature of the covalent bonding in organic compounds can be described in terms of orbitals hybridisation concept, according to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital of hydrogen to give a carbon - hydrogen (C–H) single bond (sigma, σ bond). Overlap of a sp2 orbital of one carbon with sp2 orbital of another results in the formation of a carbon–carbon σ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side- by-side) overlap to give a pi (π) bond. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula. Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds. The naming of the organic compounds is carried out by following a set of rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature, the names are correlated with the structure in such a way that the reader can deduce the structure from the name. Organic reaction mechanism concepts are based on the structure of the substrate molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects and the conditions of the reaction. These organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain carbon atom or other atom positions as places of low or high electron densities. Organic reactions can be broadly classified into following types; substitution, addition, elimination and rearrangement reactions. Purification, qualitative and quantitative analysis of organic compounds are carried out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties. Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography. Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in a pure form, its qualitative analysis is carried out for detection of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of percentages of all other elements present. Reprint 2025-26 292 chemistry Exercises 8.1 What are hybridisation states of each carbon atom in the following compounds ? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6 8.2 Indicate the σ and π bonds in the following molecules : C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3 8.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one. 8.4 Give the IUPAC names of the following compounds : (a) (b) (c) (d) (e) (f) Cl2CHCH2OH 8.5 Which of the following represents the correct IUPAC name for the compounds concer ned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne. 8.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2 8.7 Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for : (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (c) Hexanedial 8.8 Identify the functional groups in the following compounds (a) (b) (c) 8.9 Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why? 8.10 Explain why alkyl groups act as electron donors when attached to a π system. 8.11 Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. + (a) C6H5OH (b) C6H5NO2 (c) CH3CH=CHCHO (d) C6H5–CHO (e) C6H5–CH2 + (f) CH3CH=CH C H2 8.12 What are electrophiles and nucleophiles ? Explain with examples. 8.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH3COOH + HO– → CH3COO–+H2O Reprint 2025-26 organic chemistry – some basic principles and techniques 293 – (b) CH3COCH3+ C N → (CH3)2C(CN)(OH) + (c) C6H6 + CH3C O → C6H5COCH3 8.14 Classify the following reactions in one of the reaction type studied in this unit. (a) CH3CH2Br + HS– → CH3CH2SH + Br– (b) (CH3)2C = CH2 + HCI → (CH3)2CIC – CH3 (c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br– (d) (CH3)3C– CH2OH + HBr → (CH3)2CBrCH2CH2CH3 + H2O 8.15 What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ? (a) (b) (c) 8.16 For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion. (a) (b) (c) (d) 8.17 Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH (b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH 8.18 Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography 8.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S. 8.20 What is the difference between distillation, distillation under reduced pressure and steam distillation ? Reprint 2025-26 294 chemistry 8.21 Discuss the chemistry of Lassaigne’s test. 8.22 Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method. 8.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound. 8.24 Explain the principle of paper chromatography. 8.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens? 8.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. 8.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. 8.28 Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ? 8.29 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer. 8.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound? 8.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test? 8.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. 8.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. 8.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. 8.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound. 8.36 In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hydridised orbitals involved in the formation of: C2 – C3 bond is: (a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3 8.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4 8.38 Which of the following carbocation is most stable ? + + + + (a) (CH3)3C. C­H2 (b) (CH3)3C­ (c) CH3CH2C­ H2 (d) CH3C­ H CH2CH3 8.39 The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography 8.40 The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition Reprint 2025-26 Hydrocarbons 295 Unit 9 Hydrocarbons Hydrocarbons are the important sources of energy. After studying this unit, you will be able to • name hydrocarbons according to The term ‘hydrocarbon’ is self-explanatory which means IUPAC system of nomenclature; compounds of carbon and hydrogen only. Hydrocarbons • recognise and write structures play a key role in our daily life. You must be familiar o f i s o m e r s o f a l k a n e s , with the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the alkenes, alkynes and aromatic abbreviated form of liquified petroleum gas whereas CNG hydrocarbons; stands for compressed natural gas. Another term ‘LNG’ • learn about various methods of (liquified natural gas) is also in news these days. This is preparation of hydrocarbons; • distinguish between alkanes, also a fuel and is obtained by liquifaction of natural gas. alkenes, alkynes and aromatic Petrol, diesel and kerosene oil are obtained by the fractional hydrocarbons on the basis of distillation of petroleum found under the earth’s crust. physical and chemical properties; Coal gas is obtained by the destructive distillation of • draw and differentiate between coal. Natural gas is found in upper strata during drilling various conformations of ethane; of oil wells. The gas after compression is known as • a p p r e c i a t e t h e r o l e o f compressed natural gas. LPG is used as a domestic fuel hydrocarbons as sources of with the least pollution. Kerosene oil is also used as a energy and for other industrial domestic fuel but it causes some pollution. Automobiles applications; need fuels like petrol, diesel and CNG. Petrol and CNG • pr edict the for mation of the addition products of operated automobiles cause less pollution. All these fuels unsymmetrical alkenes and contain mixture of hydrocarbons, which are sources of alkynes on the basis of electronic energy. Hydrocarbons are also used for the manufacture mechanism; of polymers like polythene, polypropene, polystyrene etc. • comprehend the structure of Higher hydrocarbons are used as solvents for paints. They benzene, explain aromaticity are also used as the starting materials for manufacture and understand mechanism of many dyes and drugs. Thus, you can well understand of electrophilic substitution the importance of hydrocarbons in your daily life. In this reactions of benzene; unit, you will learn more about hydrocarbons. • predict the directive influence of substituents in monosubstituted 9.1 CLASSIFICATION benzene ring; • learn about carcinogenicity and Hydrocarbons are of different types. Depending upon toxicity. the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated Reprint 2025-26 296 chemistry (ii) unsaturated and (iii) aromatic of the general formula for alkane family hydrocarbons. Saturated hydrocarbons or homologous series? If we examine the contain carbon-carbon and carbon-hydrogen formula of different alkanes we find that single bonds. If different carbon atoms are the general formula for alkanes is CnH2n+2. It joined together to form open chain of carbon represents any particular homologue when n atoms with single bonds, they are termed is given appropriate value. Can you recall the as alkanes as you have already studied in structure of methane? According to VSEPR Unit 8. On the other hand, if carbon atoms theory (Unit 4), methane has a tetrahedral form a closed chain or a ring, they are termed structure (Fig. 9.1), in which carbon atom lies as cycloalkanes. Unsaturated hydrocarbons at the centre and the four hydrogen atoms lie contain carbon-carbon multiple bonds – at the four corners of a regular tetrahedron. double bonds, triple bonds or both. Aromatic All H-C-H bond angles are of 109.5°. hydrocarbons are a special type of cyclic compounds. You can construct a large number of models of such molecules of both types (open chain and close chain) keeping in mind that carbon is tetravalent and hydrogen is monovalent. For making models of alkanes, you can use toothpicks for bonds and plasticine balls for atoms. For alkenes, alkynes and aromatic hydrocarbons, spring models can be constructed. Fig. 9.1 Structure of methane In alkanes, tetrahedra are joined together9.2 ALKANES in which C-C and C-H bond lengths are As already mentioned, alkanes are saturated 154 pm and 112 pm respectively (Unit 8). open chain hydrocarbons containing You have already read that C–C and C–H σ carbon - carbon single bonds. Methane (CH4) bonds are formed by head-on overlapping of is the first member of this family. Methane is 3 sp hybrid orbitals of carbon and 1s orbitals a gas found in coal mines and marshy places. of hydrogen atoms. If you replace one hydrogen atom of methane by carbon and join the required number of 9.2.1 Nomenclature and Isomerism hydrogens to satisfy the tetravalence of the You have already read about nomenclature other carbon atom, what do you get? You of different classes of organic compounds get C2H6. This hydrocarbon with molecular in Unit 8. Nomenclature and isomerism formula C2H6 is known as ethane. Thus you in alkanes can further be understood with can consider C2H6 as derived from CH4 by the help of a few more examples. Common replacing one hydrogen atom by -CH3 group. names are given in parenthesis. First three Go on constructing alkanes by doing this alkanes – methane, ethane and propane have theoretical exercise i.e., replacing hydrogen only one structure but higher alkanes can atom by –CH3 group. The next molecules will have more than one structure. Let us write be C3H8, C4H10 … structures for C4H10. Four carbon atoms of H H H C4H10 can be joined either in a continuous replace any H by - CH3 chain or with a branched chain in the H—C—H H—C—C—H or C2H6 following two ways : H H H I These hydrocarbons are inert under normal conditions as they do not react with acids, bases and other reagents. Hence, they were earlier known as paraffins (latin : Butane (n- butane), (b.p. 273 K)parum, little; affinis, affinity). Can you think Reprint 2025-26 Hydrocarbons 297 II isomers. It is also clear that structures I and III have continuous chain of carbon atoms but structures II, IV and V have a branched chain. Such structural isomers which differ in chain of carbon atoms are known as chain isomers. Thus, you have seen that C4H10 2-Methylpropane (isobutane) and C5H12 have two and three chain isomers (b.p.261 K) respectively. In how many ways, you can join five Problem 9.1 carbon atoms and twelve hydrogen atoms of Write structures of different chain C5H12? They can be arranged in three ways isomers of alkanes corresponding to the as shown in structures III–V molecular formula C6H14. Also write their IUPAC names.III Solution (i) CH3 – CH2 – CH2 – CH2– CH2– CH3 n-Hexane Pentane (n-pentane) (b.p. 309 K) IV 2-Methylpentane 3-Methylpentane 2-Methylbutane (isopentane) (b.p. 301 K) 2,3-Dimethylbutane V 2,2 - Dimethylbutane Based upon the number of carbon atoms attached to a carbon atom, the carbon atom is 2,2-Dimethylpropane (neopentane) termed as primary (1°), secondary (2°), tertiary (b.p. 282.5 K) (3°) or quaternary (4°). Carbon atom attached Structures I and II possess same molecular to no other carbon atom as in methane or to formula but differ in their boiling points and only one carbon atom as in ethane is called other properties. Similarly structures III, IV primary carbon atom. Terminal carbon and V possess the same molecular formula atoms are always primary. Carbon atom but have different properties. Structures I and attached to two carbon atoms is known as II are isomers of butane, whereas structures secondary. Tertiary carbon is attached to III, IV and V are isomers of pentane. Since three carbon atoms and neo or quaternary difference in properties is due to difference in carbon is attached to four carbon atoms. Can their structures, they are known as structural you identify 1°, 2°, 3° and 4° carbon atoms in Reprint 2025-26 298 chemistry structures I to V ? If you go on constructing compounds. These groups or substituents structures for higher alkanes, you will be are known as alkyl groups as they are derived getting still larger number of isomers. C6H14 from alkanes by removal of one hydrogen has got five isomers and C7H16 has nine. As atom. General formula for alkyl groups is many as 75 isomers are possible for C10H22. CnH2n+1 (Unit 8). In structures II, IV and V, you observed Let us recall the general rules for that –CH3 group is attached to carbon atom nomenclature already discussed in Unit 8. numbered as 2. You will come across groups Nomenclature of substituted alkanes can like –CH3, –C2H5, –C3H7 etc. attached to further be understood by considering the carbon atoms in alkanes or other classes of following problem: Problem 9.2 Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain. Solution Structures of – C5H11 group Corresponding alcohols Name of alcohol (i) CH3 – CH2 – CH2 – CH2– CH2 – CH3 – CH2 – CH2 – CH2– CH2 – OH Pentan-1-ol (ii) CH3 – CH – CH2 – CH2 – CH3 CH3 – CH – CH2 – CH2– CH3 Pentan-2-ol | | OH (iii) CH3 – CH2 – CH – CH2 – CH3 CH3 – CH2 – CH – CH2– CH3 Pentan-3-ol | | OH CH3 CH3 3-Methyl- | | butan-1-ol (iv) CH3 – CH – CH2 – CH2 – CH3 – CH – CH2 – CH2– OH CH3 CH3 2-Methyl- | | butan-1-ol (v) CH3 – CH2 – CH – CH2 – CH3 – CH2 – CH – CH2– OH CH3 CH3 2-Methyl- | | butan-2-ol (vi) CH3 – C – CH2 – CH3 CH3 – C – CH2 – CH3 | | OH CH3 CH3 2,2- Dimethyl- | | propan-1-ol (vii) CH3 – C – CH2 – CH3 – C – CH2OH | | CH3 CH3 CH3 CH3 OH 3-Methyl- | | | | b u t a n - 2 - o l (viii) CH3 – CH – CH –CH3 CH3 – CH – CH –CH3 Reprint 2025-26 Hydrocarbons 299 Table 9.1 Nomenclature of a Few Organic Compounds Structure and IUPAC Name Remarks Lowest sum and alphabetical (a) 1CH3–2CH – 3CH2 – 4CH – 5CH2 – 6CH3 (4 – Ethyl – 2 – methylhexane) arrangement Lowest sum and (b) 8CH3 – 7CH2 – 6CH2 – 5CH – 4CH – 3C – 2CH2 – 1CH3 alphabetical arrangement (3,3-Diethyl-5-isopropyl-4-methyloctane) sec is not considered (c) 1CH3–2CH2–3CH2–4CH–5CH–6CH2–7CH2–8CH2–9CH2–10CH3 while arranging alphabetically; isopropyl is taken 5-sec– Butyl-4-isopropyldecane as one word (d) 1CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3 Further numbering to the substituents of the side chain 5-(2,2– Dimethylpropyl)nonane (e) 1CH3 – 2CH2 – 3CH – 4CH2 – 5CH – 6CH2 – 7CH3 Alphabetical priority order 3–Ethyl–5–methylheptane important to write the correct structure Problem 9.3 from the given IUPAC name. To do this, first Write IUPAC names of the following of all, the longest chain of carbon atoms compounds : corresponding to the parent alkane is written. (i) (CH3)3 C CH2C(CH3)3 Then after numbering it, the substituents are (ii) (CH3)2 C(C2H5)2 attached to the correct carbon atoms and (iii) tetra – tert-butylmethane finally valence of each carbon atom is satisfied by putting the correct number of hydrogen Solution atoms. This can be clarified by writing the (i) 2, 2, 4, 4-Tetramethylpentane structure of 3-ethyl-2, 2–dimethylpentane in (ii) 3, 3-Dimethylpentane the following steps : (iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - i) Draw the chain of five carbon atoms: tetramethylpentane C – C – C – C – C If it is important to write the correct ii) Give number to carbon atoms: IUPAC name for a given structure, it is equally C 1– C2– C3– C4– C5 Reprint 2025-26 300 chemistry iii) Attach ethyl group at carbon 3 and two Longest chain is of six carbon atoms and methyl groups at carbon 2 not that of five. Hence, correct name is 3-Methylhexane. 7 6 5 4 3 2 1 C1 – 2C – 3C – 4C – 5C (ii) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 iv) Satisfy the valence of each carbon atom by putting requisite number of hydrogen Numbering is to be started from the atoms : end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5- methylheptane. CH3 – C – CH – CH2 – CH3 9.2.2 Preparation Petroleum and natural gas are the main Thus we arrive at the correct structure. sources of alkanes. However, alkanes can be If you have understood writing of structure prepared by following methods : from the given name, attempt the following 1. From unsaturated hydrocarbonsproblems. Dihydrogen gas adds to alkenes and alkynes Problem 9.4 in the presence of finely divided catalysts like Write structural formulas of the following platinum, palladium or nickel to form alkanes. compounds : This process is called hydrogenation. These metals adsorb dihydrogen gas on their (i) 3, 4, 4, 5–Tetramethylheptane surfaces and activate the hydrogen – hydrogen (ii) 2,5-Dimethyhexane bond. Platinum and palladium catalyse the reaction at room temperature but relatively Solution higher temperature and pressure are required with nickel catalysts. (i) CH3 – CH2 – CH – C – CH– CH – CH2 =CH2 + H2 Pt/Pd/Ni CH3−CH3 CH3 Ethene Propane (9.1) CH2–CH=CH2 + H2 Pt/Pd/Ni CH3−CH2CH3 Propane Propane (9.2) (ii) CH3 – CH – CH2 – CH2 – CH – CH3 Problem 9.5 CH3–C≡ C–H + 2H Pt/Pd/Ni CH3−CH2CH3 Write structures for each of the Propyne Propane following compounds. Why are the given (9.3) names incorrect? Write correct IUPAC names. 2. From alkyl halides (i) 2-Ethylpentane i) Alkyl halides (except fluorides) on reduction (ii) 5-Ethyl – 3-methylheptane with zinc and dilute hydrochloric acid give Solution alkanes. (i) CH3 – CH – CH2– CH2 – CH3 Zn,H CH–C1+H2 + CH4+HC1 (9.4) Chloromethane Methane Reprint 2025-26 Hydrocarbons 301 C2H5–C1+H2 Zn,H+ C2H6+HC1 alkane containing even number of Chloroethane Ethane (9.5) carbon atoms at the anode. − + Zn,H+ 2CH3 COO Na + 2H2 OCH3CH2CH2C1 + H2 CH3CH2CH3+CH1 Sodium acetate1-Chloropropane Propane (9.6) ↓ Electrolysts ii) Alkyl halides on treatment with sodium CH3 − CH3 + 2CO2 + H2 + 2NaOH (9.9) metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction The reaction is supposed to follow the is known as Wurtz reaction and is used following path : O for the preparation of higher alkanes containing even number of carbon i) 2CH3COO–Na+ 2CH3 – C – O –+2Na+ atoms. ii) At anode: CH3Br+2Na+BrCH3 dry ether CH3+2Na O O Bromomenthane Ethane – –02e– 2C.H3+2CO2↑ (9.7) 2CH3 –C–O 2CH3 – C – dry ether Acetate ion Acetate Methyl freeC2H5Br+2Na+BrC2H5 C2H5–C2H free radical radical Bromoethane n–Butane iii) H3C + CH3 H3C–CH3↑ (9.8) iv) At cathode : What will happen if two different alkyl halides are taken? H2O+e–→–OH+ 2 →H2↑3. From carboxylic acids i) Sodium salts of carboxylic acids on Methane cannot be prepared by this heating with soda lime (mixture of sodium method. Why? hydroxide and calcium oxide) give alkanes 9.2.3 Properties containing one carbon atom less than the carboxylic acid. This process of elimination Physical properties of carbon dioxide from a carboxylic acid is Alkanes are almost non-polar molecules known as decarboxylation. because of the covalent nature of C-C and CaO C-H bonds and due to very little difference CH3COO– Na++NaOH ∆ CH4+Na2CO3 of electronegativity between carbon and Sodium ethanoate hydrogen atoms. They possess weak van der Waals forces. Due to the weak forces, the first Problem 9.6 four members, C1 to C4 are gases, C5 to C17 are liquids and those containing 18 carbon Sodium salt of which acid will be needed atoms or more are solids at 298 K. They are for the preparation of propane ? Write colourless and odourless. What do you think chemical equation for the reaction. about solubility of alkanes in water based Solution upon non-polar nature of alkanes? Petrol Butanoic acid, is a mixture of hydrocarbons and is used CH3CH2CH2COO–Na++ NaOH CaO as a fuel for automobiles. Petrol and lower fractions of petroleum are also used for dry CH3CH2CH3+Na2CO3 cleaning of clothes to remove grease stains. On the basis of this observation, what do you think about the nature of the greasy ii) Kolbe’s electrolytic method: An aqueous substance? You are correct if you say that solution of sodium or potassium salt of grease (mixture of higher alkanes) is non- a carboxylic acid on electrolysis gives Reprint 2025-26 302 chemistry polar and, hence, hydrophobic in nature. It is reducing agents. However, they undergo generally observed that in relation to solubility the following reactions under certain of substances in solvents, polar substances conditions. are soluble in polar solvents, whereas the 1. Substitution reactions non-polar ones in non-polar solvents i.e., like One or more hydrogen atoms of alkanesdissolves like. can be replaced by halogens, nitro group Boiling point (b.p.) of different alkanes are and sulphonic acid group. Halogenationgiven in Table 9.2 from which it is clear that takes place either at higher temperaturethere is a steady increase in boiling point with (573-773 K) or in the presence of diffusedincrease in molecular mass. This is due to the sunlight or ultraviolet light. Lower alkanesfact that the intermolecular van der Waals do not undergo nitration and sulphonationforces increase with increase of the molecular reactions. These reactions in which hydrogensize or the surface area of the molecule. atoms of alkanes are substituted are known You can make an interesting observation as substitution reactions. As an example,by having a look on the boiling points of chlorination of methane is given below:three isomeric pentanes viz., (pentane, 2-methylbutane and 2,2-dimethylpropane). It Halogenation is observed (Table 9.2) that pentane having a continuous chain of five carbon atoms has CH2 + C1 hv CH3C1 + HC1 the highest boiling point (309.1K) whereas Chloromethane (9.10) 2,2 – dimethylpropane boils at 282.5K. With increase in number of branched chains, hv CH3C1 + CH2 C12 + HC1 the molecule attains the shape of a sphere. Dichloromethane (9.11)This results in smaller area of contact and therefore weak intermolecular forces between CH2C12 hv CHC13 + HC1 spherical molecules, which are overcome at Trichloromethane (9.12) relatively lower temperatures. Chemical properties CHC13 + C12 hv CC14 + HC1 As already mentioned, alkanes are generally Tetrachloromethane (9.13) inert towards acids, bases, oxidising and Table 9.2 Variation of Melting Point and Boiling Point in Alkanes Molecular Name Molecular b.p./(K) m.p./(K) formula mass/u CH4 Methane 16 111.0 90.5 C2H6 Ethane 30 184.4 101.0 C3H8 Propane 44 230.9 85.3 C4H10 Butane 58 272.4 134.6 C4H10 2-Methylpropane 58 261.0 114.7 C5H12 Pentane 72 309.1 143.3 C5H12 2-Methylbutane 72 300.9 113.1 C5H12 2,2-Dimethylpropane 72 282.5 256.4 C6H14 Hexane 86 341.9 178.5 C7H16 Heptane 100 371.4 182.4 C8H18 Octane 114 398.7 216.2 C9H20 Nonane 128 423.8 222.0 C10H22 Decane 142 447.1 243.3 C20H42 Eicosane 282 615.0 309.7 Reprint 2025-26 Hydrocarbons 303 CH3–CH3 + C12 hv CH3–CH2C1 + HC1 steps are possible and may occur. Two such steps given below explain how more highly Chloroethane (9.14) haloginated products are formed. It is found that the rate of reaction of C. 1 → C. H2C1 + HC1alkanes with halogens is F2 > Cl2 > Br2 > I2. CH3C1 + Rate of replacement of hydrogens of alkanes is : C. H2C1 + C1– C1 → CH2C12 + C. 1 3° > 2° > 1°. Fluorination is too violent to (iii) Termination: The reaction stops afterbe controlled. Iodination is very slow and a some time due to consumption of reactantsreversible reaction. It can be carried out in the and / or due to the following side reactions :presence of oxidizing agents like HIO3 or HNO3. The possible chain terminating steps are: CH4+I2 CH3I+HI (9.15) (a) C. 1 + C. 1 → C1–C1 HIO3+5HI→312+3H2O (9.16) (b) H3 C. + C. H3 → H3C– CH3 Halogenation is supposed to proceed via (c) H3 C. 1 + C. 1 → H3C–C1 free radical chain mechanism involving three Though in (c), CH3 – Cl, the one of thesteps namely initiation, propagation and products is formed but free radicals aretermination as given below: consumed and the chain is terminated. The Mechanism above mechanism helps us to understand (i) Initiation : The reaction is initiated the reason for the formation of ethane as a by homolysis of chlorine molecule in the byproduct during chlorination of methane. presence of light or heat. The Cl–Cl bond 2. Combustion is weaker than the C–C and C–H bond and Alkanes on heating in the presence of air or hence, is easiest to break. dioxygen are completely oxidized to carbon dioxide and water with the evolution of large C1–C1 hv C. H3 + C1 amount of heat. homolysis Chlorine free radicals CH4 (g) + 20 2 (g) → CO2 (g) + 2H2 O(1); (ii) Propagation : Chlorine free radical Äc Hè − 890kJmol -1 attacks the methane molecule and takes the (9.17) reaction in the forward direction by breaking C 4 H10 (g) + 13/2O2 (g) → 4CO2 (g) + 5H2 O(1)the C-H bond to generate methyl free radical with the formation of H-Cl. Äc Hè = −2875.84kJmol -1 + hv + (9.18)(a) CH4 + C1 C H3 + H–C1 The general combustion equation for any The methyl radical thus obtained attacks alkane is : the second molecule of chlorine to form  3n + 1 O2 → nCO2 + (n + 1)H2 O C n H2n+2 +  CH3 – Cl with the liberation of another chlorine   2 free radical by homolysis of chlorine molecule. (9.19) hv Due to the evolution of large amount of(b) CH3 + C1–C1 CH3 – C1 + C1 heat during combustion, alkanes are used as The chlorine and methyl free radicals fuels. generated above repeat steps (a) and (b) During incomplete combustion of alkanes respectively and thereby setup a chain of with insufficient amount of air or dioxygen, reactions. The propagation steps (a) and carbon black is formed which is used in (b) are those which directly give principal the manufacture of ink, printer ink, black products, but many other propagation pigments and as filters. Reprint 2025-26 304 chemistry incomplete pressure in the presence of oxides of vanadium,CH4(g) + O2(g) combustion C(s)+2H2 O(1) molybdenum or chromium supported over (9.20) alumina get dehydrogenated and cyclised to 3. Controlled oxidation benzene and its homologues. This reaction is Alkanes on heating with a regulated supply known as aromatization or reforming. of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products. (i) 2CH4 + O2 Cu/523K/100atm 2CH3 OH Methanol (9.21) Mo2O3 (9.26)(ii) CH4 + O2 HCHO + H2O ∆ Toluene (C7H8) is methyl derivative of Methanal benzene. Which alkane do you suggest for preparation of toluene ? (9.22) 6. Reaction with steam (iii) 2CH3CH3 + 3O2 (CH3COO)Mn 2CH3COOH ∆ Methane reacts with steam at 1273 K in the Ethanoic acid presence of nickel catalyst to form carbon + 2H2O monoxide and dihydrogen. This method is (9.23) used for industrial preparation of dihydrogen (iv) Ordinarily alkanes resist oxidation but gas alkanes having tertiary H atom can be Ni oxidized to corresponding alcohols by CH4 + H2IO ∆ CO + 3H2 (9.27) potassium permanganate. 7. Pyrolysis KMnO4 (iCH3)3 CH Oxidation (CH3)3 COH Higher alkanes on heating to higher temperature decompose into lower alkanes, 2-Methylpropane 2-Methylpropane-2-01 alkenes etc. Such a decomposition reaction (9.24) into smaller fragments by the application of heat is called pyrolysis or cracking.4. Isomerisation n-Alkanes on heating in the presence of anhydrous aluminium chloride and hydrogen chloride gas isomerise to branched chain alkanes. Major products are given below. Some minor products are also possible which you can think over. Minor products are (9.28) generally not reported in organic reactions. Pyrolysis of alkanes is believed to be a Anhy, AICI3/ HCI free radical reaction. Preparation of oil gasCH3(CH)2)4CH3 or petrol gas from kerosene oil or petrol n-Hexane involves the principle of pyrolysis. For CH3CH–(CH2)2–CH3+CH3CH2–CH–CH2–CH3 example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of CH3 CH3 platinum, palladium or nickel gives a mixture 2-Methylpentane 3-Methylpenatone of heptane and pentene. (9.25) Pt/Pd/Ni Other 5. Aromatization C12H26 973K C7H16 + C5H10 + Products n-Alkanes having six or more carbon atoms Dodecane Heptane Pentene on heating to 773K at 10-20 atmospheric (9.29) Reprint 2025-26 Hydrocarbons 305 9.2.4 Conformations 1. Sawhorse projections Alkanes contain carbon-carbon sigma (σ) In this projection, the molecule is viewed bonds. Electron distribution of the sigma along the molecular axis. It is then projected molecular orbital is symmetrical around the on paper by drawing the central C–C bond internuclear axis of the C–C bond which is as a somewhat longer straight line. Upper not disturbed due to rotation about its axis. end of the line is slightly tilted towards This permits free rotation about C–C single right or left hand side. The front carbon is bond. This rotation results into different shown at the lower end of the line, whereas spatial arrangements of atoms in space the rear carbon is shown at the upper end. which can change into one another. Such Each carbon has three lines attached to it spatial arrangements of atoms which can corresponding to three hydrogen atoms. be converted into one another by rotation The lines are inclined at an angle of 120° to around a C-C single bond are called each other. Sawhorse projections of eclipsed conformations or conformers or rotamers. and staggered conformations of ethane are Alkanes can thus have infinite number of depicted in Fig. 9.2. conformations by rotation around C-C single bonds. However, it may be remembered that rotation around a C-C single bond is not completely free. It is hindered by a small energy barrier of 1-20 kJ mol–1 due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Conformations of ethane : Ethane Fig. 9.2 Sawhorse projections of ethane molecule (C2H6) contains a carbon – carbon 2. Newman projectionssingle bond with each carbon atom attached In this projection, the molecule is viewed at theto three hydrogen atoms. Considering the C–C bond head on. The carbon atom nearerball and stick model of ethane, keep one to the eye is represented by a point. Threecarbon atom stationary and rotate the other carbon atom around the C-C axis. This hydrogen atoms attached to the front carbon rotation results into infinite number of spatial atom are shown by three lines drawn at an arrangements of hydrogen atoms attached to angle of 120° to each other. The rear carbon one carbon atom with respect to the hydrogen atom (the carbon atom away from the eye) is atoms attached to the other carbon atom. represented by a circle and the three hydrogen These are called conformational isomers atoms are shown attached to it by the shorter (conformers). Thus there are infinite number lines drawn at an angle of 120° to each other. of conformations of ethane. However, there are The Newman’s projections are depicted in two extreme cases. One such conformation Fig. 9.3. in which hydrogen atoms attached to two carbons are as closed together as possible is called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the staggered conformation. Any other intermediate conformation is called a skew conformation.It may be remembered that in all the conformations, the bond angles and the bond lengths remain the same. Eclipsed and the staggered conformations can be represented by Sawhorse and Newman projections. Fig. 9.3 Newman’s projections of ethane Reprint 2025-26 306 chemistry Relative stability of conformations: As the first member, ethylene or ethene (C2H4) mentioned earlier, in staggered form of ethane, was found to form an oily liquid on reaction the electron clouds of carbon-hydrogen bonds with chlorine. are as far apart as possible. Thus, there are 9.3.1 Structure of Double Bond minimum repulsive forces, minimum energy Carbon-carbon double bond in alkenesand maximum stability of the molecule. On the consists of one strong sigma (σ) bond (bondother hand, when the staggered form changes enthalpy about 397 kJ mol –1) due to head-oninto the eclipsed form, the electron clouds of the 2 overlapping of sp hybridised orbitals andcarbon – hydrogen bonds come closer to each one weak pi (π) bond (bond enthalpy aboutother resulting in increase in electron cloud 284 kJ mol –1) obtained by lateral or sidewaysrepulsions. To check the increased repulsive overlapping of the two 2p orbitals of the twoforces, molecule will have to possess more carbon atoms. The double bond is shorter inenergy and thus has lesser stability. As already bond length (134 pm) than the C–C singlementioned, the repulsive interaction between bond (154 pm). You have already read thatthe electron clouds, which affects stability of the pi (π) bond is a weaker bond due to poora conformation, is called torsional strain. sideways overlapping between the two 2pMagnitude of torsional strain depends upon orbitals. Thus, the presence of the pi (π) bondthe angle of rotation about C–C bond. This makes alkenes behave as sources of looselyangle is also called dihedral angle or torsional held mobile electrons. Therefore, alkenes areangle. Of all the conformations of ethane, the easily attacked by reagents or compoundsstaggered form has the least torsional strain which are in search of electrons. Suchand the eclipsed form, the maximum torsional reagents are called electrophilic reagents.strain. Therefore, staggered conformation is The presence of weaker π-bond makes alkenesmore stable than the eclipsed conformation. unstable molecules in comparison to alkanesHence, molecule largely remains in staggered and thus, alkenes can be changed into singleconformation or we can say that it is preferred bond compounds by combining with theconformation. Thus it may be inferred that electrophilic reagents. Strength of the doublerotation around C–C bond in ethane is not bond (bond enthalpy, 681 kJ mol–1) is greatercompletely free. The energy difference between than that of a carbon-carbon single bond inthe two extreme forms is of the order of 12.5 ethane (bond enthalpy, 348 kJ mol –1). OrbitalkJ mol–1, which is very small. Even at ordinary diagrams of ethene molecule are shown in temperatures, the ethane molecule gains Figs. 9.4 and 9.5. thermal or kinetic energy sufficient enough to overcome this energy barrier of 12.5 kJ mol–1 through intermolecular collisions. Thus, it can be said that rotation about carbon-carbon single bond in ethane is almost free for all practical purposes. It has not been possible to separate and isolate different conformational isomers of ethane.

8.3 Draw the structures of the following compounds. (i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid

8.4 Classification of Organic (homocyclic). Compounds The existing large number of organic compounds and their ever -increasing numbers has made it necessary to classify them on the basis of their structures. Organic Cyclopropane Cyclohexane Cyclohexene compounds are broadly classified as follows: Sometimes atoms other than carbon are also present in the ring (heterocylic). Tetrahydrofuran given below is an example of this type of compound: Tetrahydrofuran These exhibit some of the properties similar to those of aliphatic compounds. (b) Aromatic compounds Aromatic compounds are special types of compounds. You will learn about these compounds in detail in Unit 9. These include benzene and other related ring compounds (benzenoid). Like alicyclic compounds, aromatic comounds may also have hetero atom in the ring. Such compounds are called I. Acyclic or open chain compounds hetrocyclic aromatic compounds. Some of the examples of various types of aromatic These compounds are also called as aliphatic compounds are: compounds and consist of straight or branched chain compounds, for example: Benzenoid aromatic compounds CH3CH3 Ethane Isobutane Benzene Aniline Naphthalene Non-benzenoid compound Acetaldehyde Acetic acid II Cyclic or closed chain or ring compounds (a) Alicyclic compounds Tropone Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring Reprint 2025-26 262 chemistry Heterocyclic aromatic compounds so because it is found in citrus fruits and the acid found in red ant is named formic acid since the Latin word for ant is formica. These names are traditional and are considered as trivial or common names. Some common Furan Thiophene Pyridine names are followed even today. For example, Organic compounds can also be classified Buckminsterfullerene is a common name on the basis of functional groups, into families given to the newly discovered C60 cluster (a or homologous series. form of carbon) noting its structural similarity to the geodesic domes popularised by the8.4.1 Functional Group famous architect R. Buckminster Fuller. The functional group is an atom or a group Common names are useful and in many of atoms joined to the carbon chain which is cases indispensable, particularly when the responsible for the characteristic chemical alternative systematic names are lengthy and properties of the organic compounds. The complicated. Common names of some organic examples are hydroxyl group (–OH), aldehyde compounds are given in Table 8.1. group (–CHO) and carboxylic acid group (– COOH) etc. Table 8.1 Common or Trivial Names of Some Organic Compounds 8.4.2 Homologous Series A group or a series of organic compounds each containing a characteristic functional group forms a homologous series and the members of the series are called homologues. The members of a homologous series can be represented by general molecular formula and the successive members differ from each other in molecular formula by a –Ch2 unit. There are a number of homologous series of organic compounds. Some of these are alkanes, alkenes, alkynes, haloalkanes, alkanols, alkanals, alkanones, alkanoic acids, amines etc. It is also possible that a compound contains two or more identical or different functional groups. This gives rise to polyfunctional compounds.

8.19 An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

8.5 Draw structures of the following derivatives. (i) The 2,4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde 255 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26

8.1 What is meant by the following terms ? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base

8.5 NOMENCLATURE OF ORGANIC COMPOUNDS Organic chemistry deals with millions of 8.5.1 The IUPAC System of Nomenclature compounds. In order to clearly identify A systematic name of an organic compound them, a systematic method of naming has is generally derived by identifying the parent been developed and is known as the IUPAC hydrocarbon and the functional group(s)(International Union of Pure and Applied attached to it. See the example given below.Chemistry) system of nomenclature. In this systematic nomenclature, the names are correlated with the structure such that the reader or listener can deduce the structure from the name. Before the IUPAC system of nomenclature, however, organic compounds were assigned names based on their origin or certain properties. For instance, citric acid is named Reprint 2025-26 organic chemistry – some basic principles and techniques 263 By further using prefixes and suffixes, the In order to name such compounds, the names parent name can be modified to obtain the of alkyl groups are prefixed to the name of actual name. Compounds containing carbon parent alkane. An alkyl group is derived and hydrogen only are called hydrocarbons. from a saturated hydrocarbon by removing A hydrocarbon is termed saturated if it a hydrogen atom from carbon. Thus, CH4 contains only carbon-carbon single bonds. becomes -CH3 and is called methyl group. An The IUPAC name for a homologous series of alkyl group is named by substituting ‘yl’ for such compounds is alkane. Paraffin (Latin: ‘ane’ in the corresponding alkane. Some alkyl little affinity) was the earlier name given to groups are listed in Table 8.3. these compounds. Unsaturated hydrocarbons Table 8.3 Some Alkyl Groups are those, which contain at least one carbon- carbon double or triple bond. 8.5.2 IUPAC Nomenclature of Alkanes Straight chain hydrocarbons: The names of such compounds are based on their chain structure, and end with suffix ‘-ane’ and carry a prefix indicating the number of carbon atoms present in the chain (except from CH4 to C4H10, where the prefixes are derived from trivial names). The IUPAC names of some straight chain saturated hydrocarbons are Abbreviations are used for some alkylgiven in Table 8.2. The alkanes in Table groups. For example, methyl is abbreviated as 8.2 differ from each other by merely the Me, ethyl as Et, propyl as Pr and butyl as Bu. number of -CH2 groups in the chain. They are The alkyl groups can be branched also. Thus, homologues of alkane series. propyl and butyl groups can have branched Table 8.2 IUPAC Names of Some Unbranched structures as shown below. Saturated Hydrocarbons CH3-CH- CH3-CH2-CH- CH3-CH-CH2-    CH3 CH3 CH3 Isopropyl- sec-Butyl- Isobutyl- CH3 CH3   CH3-C- CH3-C-CH2-   CH3 CH3 tert-Butyl- Neopentyl- Branched chain hydrocarbons: In a Common branched groups have specific branched chain compound small chains of trivial names. For example, the propyl groups carbon atoms are attached at one or more can either be n-propyl group or isopropyl carbon atoms of the parent chain. The small group. The branched butyl groups are called carbon chains (branches) are called alkyl sec-butyl, isobutyl and tert-butyl group. groups. For example: We also encounter the structural unit, –CH2C(CH3)3, which is called neopentyl group. CH3–CH–CH2–CH3 CH3–CH–CH2–CH–CH3    Nomenclature of branched chain alkanes: We encounter a number of branched chain CH3 CH2CH3 CH3 alkanes. The rules for naming them are given (a) (b) below. Reprint 2025-26 264 chemistry 1. First of all, the longest carbon chain in separated from the groups by hyphens the molecule is identified. In the example and there is no break between methyl (I) given below, the longest chain has nine and nonane.] carbons and it is considered as the parent 4. If two or more identical substituent or root chain. Selection of parent chain as groups are present then the numbers shown in (II) is not correct because it has are separated by commas. The names of only eight carbons. identical substituents are not repeated, instead prefixes such as di (for 2), tri (for 3), tetra (for 4), penta (for 5), hexa (for 6) etc. are used. While writing the name of the substituents in alphabetical order, these prefixes, however, are not considered. Thus, the following compounds are named as: CH3 CH3 CH3 CH3     CH3-CH-CH2-CH-CH3 CH3 C CH2CH CH3 1 2 3 4 5 1 2 3 4 5 2,4-Dimethylpentane 2,2,4-Trimethylpentane2. The carbon atoms of the parent chain are numbered to identify the parent alkane H3C H2C CH3 and to locate the positions of the carbon   atoms at which branching takes place due CH3CH2CHCCH2CH2CH3 to the substitution of alkyl group in place 1 2 3 4 5 6 7 of hydrogen atoms. The numbering is CH3 done in such a way that the branched 3-Ethyl-4,4-dimethylheptane carbon atoms get the lowest possible numbers. Thus, the numbering in the 5. If the two substituents are found in above example should be from left to right equivalent positions, the lower number (branching at carbon atoms 2 and 6) and is given to the one coming first in the not from right to left (giving numbers alphabetical listing. Thus, the following 4 and 8 to the carbon atoms at which compound is 3-ethyl-6-methyloctane and branches are attached). not 6-ethyl-3-methyloctane. 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 C  C  C  C  C  C  C  C  C   CH3 — CH2—CH—CH2—CH2—CH—CH2 —CH3   C CC CH2CH3 CH3 9 8 7 6 5 4 3 2 1 C  C  C  C  C  C  C  C  C 6. The branched alkyl groups can be   named by following the above mentioned C C  C procedures. However, the carbon atom 3. The names of alkyl groups attached of the branch that attaches to the root as a branch are then prefixed to the alkane is numbered 1 as exemplified name of the parent alkane and position below. of the substituents is indicated by the 4 3 2 1 appropriate numbers. If different alkyl CH3–CH–CH2–CH– groups are present, they are listed in alphabetical order. Thus, name for the   compound shown above is: 6-ethyl-2- CH3 CH3 1,3-Dimethylbutyl- methylnonane. [Note: the numbers are Reprint 2025-26 organic chemistry – some basic principles and techniques 265 The name of such branched chain alkyl group Cyclic Compounds: A saturated monocyclic is placed in parenthesis while naming the compound is named by prefixing ‘cyclo’ to the compound. While writing the trivial names corresponding straight chain alkane. If side of substituents’ in alphabetical order, the chains are present, then the rules given above prefixes iso- and neo- are considered to be are applied. Names of some cyclic compounds the part of the fundamental name of alkyl are given below. group. The prefixes sec- and tert- are not considered to be the part of the fundamental name. The use of iso and related common prefixes for naming alkyl groups is also allowed by the IUPAC nomenclature as long as these are not further substituted. In multi- substituted compounds, the following rules may aso be remembered: • If there happens to be two chains of equal size, then that chain is to be selected which contains more number of side chains. 3-Ethyl-1,1-dimethylcyclohexane • After selection of the chain, numbering (not 1-ethyl-3,3-dimethylcyclohexane) is to be done from the end closer to the substituent. Problem 8.7 Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect. 2,5,6- Trimethyloctane [and not 3,4,7-Trimethyloctane] 5-(2-Ethylbutyl)-3,3-dimethyldecane [and not 5-(2,2-Dimethylbutyl)-3-ethyldecane] 3-Ethyl-5-methylheptane [and not 5-Ethyl-3-methylheptane] Solution (a) Lowest locant number, 2,5,6 is lower than 3,5,7, (b) substituents are 5-sec-Butyl-4-isopropyldecane in equivalent position; lower number is given to the one that comes first in the name according to alphabetical order. 8.5.3 N o m e n c l a t u r e o f O r g a n i c Compounds having Functional Group(s) A functional group, as defined earlier, is an atom or a group of atoms bonded together in 5-(2,2-Dimethylpropyl)nonane a unique manner which is usually the site of Reprint 2025-26 266 chemistry chemical reactivity in an organic molecule. class suffix. In such cases the full name of the Compounds having the same functional parent alkane is written before the class suffix. group undergo similar reactions. For example, For example CH2(OH)CH2(OH) is named as CH3OH, CH3CH2OH, and (CH3)2CHOH — ethane–1,2–diol. However, the ending – ne of all having -OH functional group liberate the parent alkane is dropped in the case of hydrogen on reaction with sodium metal. compounds having more than one double or The presence of functional groups enables triple bond; for example, CH2=CH-CH=CH2 is systematisation of organic compounds into named as buta–1,3–diene. different classes. Examples of some functional groups with their prefixes and suffixes along Problem 8.8 with some examples of organic compounds Write the IUPAC names of the compounds possessing these are given in Table 8.4. i-iv from their given structures. First of all, the functional group present in the molecule is identified which determines the choice of appropriate suffix. The longest chain of carbon atoms containing the functional group is numbered in such a way Solutionthat the functional group is attached at the carbon atom possessing lowest possible • The functional group present is an number in the chain. By using the suffix as alcohol (OH). Hence the suffix is ‘-ol’. given in Table 8.4, the name of the compound • The longest chain containing -OH is arrived at. has eight carbon atoms. Hence the In the case of polyfunctional compounds, corresponding saturated hydrocar- one of the functional groups is chosen as the bon is octane. principal functional group and the compound • The -OH is on carbon atom 3. In is then named on that basis. The remaining addition, a methyl group is attached functional groups, which are subordinate at 6th carbon. functional groups, are named as substituents Hence, the systematic name of this using the appropriate prefixes. The choice of compound is 6-Methyloctan-3-ol. principal functional group is made on the basis of order of preference. The order of decreasing priority for some functional groups is: -COOH, –SO3H, -COOR (R=alkyl group), COCl, -CONH2, -CN,-HC=O, >C=O, -OH, -NH2, > C=C<, -C≡C- . Solution The –R, C6H5-, halogens (F, Cl, Br, I), –NO2, The functional group present is ketone alkoxy (–OR) etc. are always prefix (>C=O), hence suffix ‘-one’. Presence substituents. Thus, a compound containing of two keto groups is indicated by ‘di’, both an alcohol and a keto group is named hence suffix becomes ‘dione’. The two as hydroxyalkanone since the keto group is keto groups are at carbons 2 and 4. preferred to the hydroxyl group. The longest chain contains 6 carbon For example, HOCH2(CH2)3CH2COCH3 will atoms, hence, parent hydrocarbon is be named as 7-hydroxyheptan-2-one and not hexane. Thus, the systematic name is as 2-oxoheptan -7-ol. Similarly, BrCH2CH=CH2 Hexane-2,4-dione. is named as 3-bromoprop-1-ene and not 1-bromoprop-2-ene. If more than one functional group of the same type are present, their number is indicated by adding di, tri, etc. before the Reprint 2025-26 organic chemistry – some basic principles and techniques 267 Table 8.4 Some Functional Groups and Classes of Organic Compounds Reprint 2025-26 268 chemistry Solution (iii) Six membered ring containing a carbon-carbon double bond is implied Here, two functional groups namely by cyclohexene, which is numbered as ketone and carboxylic acid are present. shown in (I). The prefix 3-nitro means The principal functional group is the that a nitro group is present on C-3. carboxylic acid group; hence the parent Thus, complete structural formula of the chain will be suffixed with ‘oic’ acid. compound is (II). Double bond is suffixed Numbering of the chain starts from functional group whereas NO2 is prefixed carbon of – COOH functional group. functional group therefore double bond The keto group in the chain at carbon gets preference over –NO2 group: 5 is indicated by ‘oxo’. The longest chain including the principal functional group has 6 carbon atoms; hence the parent hydrocarbon is hexane. The compound is, therefore, named as 5-Oxohexanoic acid. (iv) ‘1-ol’ means that a -OH group is Solution present at C-1. OH is suffixed functional The two C=C functional groups are group and gets preference over C=C present at carbon atoms 1 and 3, while bond. Thus the structure is as shown the C≡C functional group is present at in (II): carbon 5. These groups are indicated by suffixes ‘diene’ and ‘yne’ respectively. The longest chain containing the functional groups has 6 carbon atoms; hence the parent hydrocarbon is hexane. The name of compound, therefore, is Hexa-1,3- dien-5-yne. (v) ‘heptanal’ indicates the compound Problem 8.9 to be an aldehyde containing 7 carbon Derive the structure of (i) 2-Chlorohexane, atoms in the parent chain. The (ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene, ‘6-hydroxy’ indicates that -OH group is (iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxy- present at carbon 6. Thus, the structural heptanal. formula of the compound is: CH3CH(OH) Solution CH2CH2CH2CH2CHO. Carbon atom of – CHO group is included while numbering (i) ‘hexane’ indicates the presence of the carbon chain. 6 carbon atoms in the chain. The functional group chloro is present at carbon 2. Hence, the structure of the 8.5.4 Nomenclature of Substituted compound is CH3CH2CH2CH2CH(Cl)CH3. Benzene Compounds (ii) ‘pent’ indicates that parent For IUPAC nomenclature of substituted hydrocarbon contains 5 carbon atoms benzene compounds, the substituent is in the chain. ‘en’ and ‘ol’ correspond to placed as prefix to the word benzene as shown the functional groups C=C and -OH at in the following examples. However, common carbon atoms 4 and 2 respectively. Thus, names (written in bracket below) of many the structure is substituted benzene compounds are also CH2=CHCH2CH (OH)CH3. universally used. Reprint 2025-26 organic chemistry – some basic principles and techniques 269 Substituent of the base compound is assigned number1 and then the direction of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in Methylbenzene Methoxybenzene Aminobenzene alphabetical order. Some examples are given (Toluene) (Anisole) (Aniline) below. Nitrobenzene Bromobenzene 1-Chloro-2,4-dinitrobenzene (not 4-chloro,1,3-dinitrobenzene) If benzene ring is disubstituted, the position of substituents is defined by numbering the carbon atoms of the ring such that the substituents are located at the lowest numbers possible. For example, the compound(b) is named as 1,3-dibromobenzene and not as 1,5-dibromobenzene. 2-Chloro-1-methyl-4-nitrobenzene (not 4-methyl-5-chloro-nitrobenzene) (a) (b) (c) 1,2-Dibromo- 1,3-Dibromo- 1,4-Dibromo- benzene benzene benzene In the trivial system of nomenclature 2-Chloro-4-methylanisole 4-Ethyl-2-methylanilinethe terms ortho (o), meta (m) and para (p) are used as prefixes to indicate the relative positions 1,2;1,3 and 1,4 respectively. Thus, 1,3-dibromobenzene (b) is named as m-dibromobenzene (meta is abbreviated as m-) and the other isomers of dibromobenzene 1,2-(a) and 1,4-(c), are named as ortho (or just o-) and para (or just p-)-dibromobenzene, respectively. For tri - or higher substituted benzene 3,4-Dimethylphenol derivatives, these prefixes cannot be used and the compounds are named by identifying When a benzene ring is attached to an substituent positions on the ring by following alkane with a functional group, it is considered the lowest locant rule. In some cases, common as substituent, instead of a parent. The name name of benzene derivatives is taken as the for benzene as substituent is phenyl (C6H5-, base compound. also abbreviated as Ph). Reprint 2025-26 270 chemistry different carbon skeletons, these are referred Problem 8.10 to as chain isomers and the phenomenon is Write the structural formula of: termed as chain isomerism. For example, (a) o-Ethylanisole, (b) p-Nitroaniline, C5H12 represents three compounds: (c) 2,3 - Dibromo -1 - phenylpentane, (d) 4-Ethyl-1-fluoro-2-nitrobenzene. CH3  Solution CH3CH2CH2CH2CH3 CH3−CHCH2CH3 Pentane Isopentane (2-Methylbutane) CH3  CH3 C CH3 (a) (b)  CH3 Neopentane (2,2-Dimethylpropane) (ii) Position isomerism: When two or more compounds differ in the position of (c) (d) substituent atom or functional group on the carbon skeleton, they are called position

8.18 Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclo- hexanone does not. (ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

7.21 Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol to 2,4,6-tribromophenol. (iv) Benzyl alcohol to benzoic acid. (v) Dehydration of propan-2-ol to propene. (vi) Butan-2-one to butan-2-ol.

7.7 Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol

7.17 Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO4 solution. (ii) Bromine in CS2 with phenol. (iii) Dilute HNO3 with phenol. (iv) Treating phenol wih chloroform in presence of aqueous NaOH.

7.8 Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

7.16 Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

7.6 Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl –ZnCl2 (b) HBr and (c) SOCl2. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol

7.12 Predict the products of the following reactions: (i) CH3 − CH2 − CH2 − O – CH3 + HBr → (ii) (iii) (iv) ( CH 3 ) 3 C − OC 2 H5 HI → SummarySummarySummarySummarySummary Alcohols and phenols are classified (i) on the basis of the number of hydroxyl groups and (ii) according to the hybridisation of the carbon atom, sp 3 or sp 2 to which the –OH group is attached. Ethers are classified on the basis of groups attached to the oxygen atom. Alcohols may be prepared (1) by hydration of alkenes (i) in presence of an acid and (ii) by hydroboration-oxidation reaction (2) from carbonyl compounds by (i) catalytic reduction and (ii) the action of Grignard reagents. Phenols may be prepared by (1) substitution of (i) halogen atom in haloarenes and (ii) sulphonic acid group in aryl sulphonic acids, by –OH group (2) by hydrolysis of diazonium salts and (3) industrially from cumene. Alcohols are higher boiling than other classes of compounds, namely hydrocarbons, ethers and haloalkanes of comparable molecular masses. The ability of alcohols, phenols and ethers to form intermolecular hydrogen bonding with water makes them soluble in it. Alcohols and phenols are acidic in nature. Electron withdrawing groups in phenol increase its acidic strength and electron releasing groups decrease it. Alcohols undergo nucleophilic substitution with hydrogen halides to yield alkyl halides. Dehydration of alcohols gives alkenes. On oxidation, primary alcohols yield aldehydes with mild oxidising agents and carboxylic acids with strong oxidising agents while secondary alcohols yield ketones. Tertiary alcohols are resistant to oxidation. The presence of –OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to ortho and para positions due to resonance effect. Reimer-Tiemann reaction of phenol yields salicylaldehyde. In presence of sodium hydroxide, phenol generates phenoxide ion which is even more reactive than phenol. Thus, in alkaline medium, phenol undergoes Kolbe’s reaction. Ethers may be prepared by (i) dehydration of alcohols and (ii) Williamson synthesis. The boiling points of ethers resemble those of alkanes while their solubility is comparable to those of alcohols having same molecular mass. The C–O bond in ethers can be cleaved by hydrogen halides. In electrophilic substitution, the alkoxy group activates the aromatic ring and directs the incoming group to ortho and para positions. 221 Alcohols, Phenols and Ethers Reprint 2025-26 Exercises 7.1 Write IUPAC names of the following compounds: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) C6H5–O–C2H5 (xi) C6H5–O–C7H15(n–) (xii) 7.2 Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane –1, 3, 5-triol (iv) 2,3 – Diethylphenol (v) 1 – Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 4-Chloro-3-ethylbutan-1-ol. 7.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols. 7.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane? 7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact. 7.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example. 7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O. 7.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason. 7.9 Give the equations of reactions for the preparation of phenol from cumene. 7.10 Write chemical reaction for the preparation of phenol from chlorobenzene. 7.11 Write the mechanism of hydration of ethene to yield ethanol. 7.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents. Chemistry 222 Reprint 2025-26

7.13 Show how will you synthesise: (i) 1-phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction. (iii) pentan-1-ol using a suitable alkyl halide?

7.23 Give IUPAC names of the following ethers: 7.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane 7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers. 7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction. 7.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. 7.28 Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether. 7.29 Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring. 7.30 Write the mechanism of the reaction of HI with methoxymethane. 7.31 Write equations of the following reactions: (i) Friedel-Crafts reaction – alkylation of anisole. (ii) Nitration of anisole. (iii) Bromination of anisole in ethanoic acid medium. (iv) Friedel-Craft’s acetylation of anisole. 7.32 Show how would you synthesise the following alcohols from appropriate alkenes? CH3 OH (i) OH (ii) OH (iii) (iv) OH 7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. Chemistry 224 Reprint 2025-26 Answers to Some Intext Questions 7.1 Primary alcohols (i), (ii), (iii) Secondary alcohols (iv) and (v) Tertiary alcohols (vi) 7.2 Allylic alcohols (ii) and (vi) 7.3 (i) 4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol (ii) 2, 5-Dimethylhexane-1,3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol 7.4 OH CH2 C OCH3 7.5 (i) CH3 CH CH3 (ii) O OH (iii) CH3 CH2 CH CH2 OH CH3 7.7 (i) 1-Methylcyclohexene (ii) A Mixture of but-1-ene and but-2-ene. But-2-ene is the major product formed due to rearrangement to give secondary carbocation.

7.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

7.20 How are the following conversions carried out? (i) Propene ® Propan-2-ol. (ii) Benzyl chloride ® Benzyl alcohol. (iii) Ethyl magnesium chloride ® Propan-1-ol. (iv) Methyl magnesium bromide ® 2-Methylpropan-2-ol.

7.11 (ii) 7.12 (i) CH3 CH2 CH2 OH  CH 3 Br (ii) (iii) (iv)  CH 3 3 C  I  C 2 H 5 OH Chemistry 226 Reprint 2025-26 UnitUnitUnitUnit Unit88 Objectives AldehydesAldehydesAldehydes,AldehydesAldehydesAldehydesAldehydesAldehydes,AldehydesAldehydes KetonesKetonestonestonestonestonestonestonestonestones After studying this Unit, you will be able to andandandandandandandandandand CarboxylicCarboxylicCarboxylicCarboxylicCarboxylicCarboxylicCarboxylicCarboxylicCarboxylicCarboxylic• write the common and IUPAC names of aldehydes, ketones and carboxylic acids; AcidsAcidscidscidscidscidscidscidscidscids• write the structures of the compounds containing functional groups namely carbonyl and Carbonyl compounds are of utmost importance to organic carboxyl groups; chemistry. They are constituents of fabrics, flavourings, plastics • describe the important methods and drugs. of preparation and reactions of these classes of compounds; In the previous Unit, you have studied organic • correlate physical properties and compounds with functional groups containing carbon- chemical reactions of aldehydes, oxygen single bond. In this Unit, we will study about the ketones and carboxylic acids, organic compounds containing carbon-oxygen double with their structures; bond (>C=O) called carbonyl group, which is one of the • explain the mechanism of a few most important functional groups in organic chemistry. selected reactions of aldehydes In aldehydes, the carbonyl group is bonded to a and ketones; carbon and hydrogen while in the ketones, it is bonded • understand various factors to two carbon atoms. The carbonyl compounds in which affecting the acidity of carboxylic carbon of carbonyl group is bonded to carbon or acids and their reactions; hydrogen and oxygen of hydroxyl moiety (-OH) are • describe the uses of aldehydes, known as carboxylic acids, while in compounds where ketones and carboxylic acids. carbon is attached to carbon or hydrogen and nitrogen of -NH2 moiety or to halogens are called amides and acyl halides respectively. Esters and anhydrides are derivatives of carboxylic acids. The general formulas of these classes of compounds are given below: Reprint 2025-26 Aldehydes, ketones and carboxylic acids are widespread in plants and animal kingdom. They play an important role in biochemical processes of life. They add fragrance and flavour to nature, for example, vanillin (from vanilla beans), salicylaldehyde (from meadow sweet) and cinnamaldehyde (from cinnamon) have very pleasant fragrances. They are used in many food products and pharmaceuticals to add flavours. Some of these families are manufactured for use as solvents (i.e., acetone) and for preparing materials like adhesives, paints, resins, perfumes, plastics, fabrics, etc. 8.18.18.18.18.1 NomenclatureNomenclatureNomenclatureNomenclatureNomenclature andandandandand StructureStructureStructureStructureStructure ofofofofof CarbonylCarbonylCarbonylCarbonylCarbonyl GroupGroupGroupGroupGroup 8.1.1 I. Aldehydes and ketones Nomenclature Aldehydes and ketones are the simplest and most important carbonyl compounds. There are two systems of nomenclature of aldehydes and ketones. (a) Common names Aldehydes and ketones are often called by their common names instead of IUPAC names. The common names of most aldehydes are derived from the common names of the corresponding carboxylic acids [Section 8.6.1] by replacing the ending –ic of acid with aldehyde. At the same time, the names reflect the Latin or Greek term for the original source of the acid or aldehyde. The location of the substituent in the carbon chain is indicated by Greek letters a, b, g, d, etc. The a-carbon being the one directly linked to the aldehyde group, b- carbon the next, and so on. For example Chemistry 228 Reprint 2025-26 The common names of ketones are derived by naming two alkyl or aryl groups bonded to the carbonyl group. The locations of substituents are indicated by Greek letters, a a¢, b b¢ and so on beginning with the carbon atoms next to the carbonyl group, indicated as aa¢. Some ketones have historical common names, the simplest dimethyl ketone is called acetone. Alkyl phenyl ketones are usually named by adding the name of acyl group as prefix to the word phenone. For example (b) IUPAC names The IUPAC names of open chain aliphatic aldehydes and ketones are derived from the names of the corresponding alkanes by replacing the ending –e with –al and –one respectively. In case of aldehydes the longest carbon chain is numbered starting from the carbon of the aldehyde group while in case of ketones the numbering begins from the end nearer to the carbonyl group. The substituents are prefixed in alphabetical order along with numerals indicating their positions in the carbon chain. The same applies to cyclic ketones, where the carbonyl carbon is numbered one. When the aldehyde group is attached to a ring, the suffix carbaldehyde is added after the full name of the cycloalkane. The numbering of the ring carbon atoms start from the carbon atom attached to the aldehyde group. The name of the simplest aromatic aldehyde carrying the aldehyde group on a benzene ring is benzenecarbaldehyde. However, the common name benzaldehyde is also accepted by IUPAC. Other aromatic aldehydes are hence named as substituted benzaldehydes. 229 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 The common and IUPAC names of some aldehydes and ketones are given in Table 8.1. Table 8.1: Common and IUPAC Names of Some Aldehydes and Ketones Structure Common name IUPAC name Aldehydes HCHO Formaldehyde Methanal CH3CHO Acetaldehyde Ethanal (CH3)2CHCHO Isobutyraldehyde 2-Methylpropanal g-Methylcyclohexanecarbaldehyde 3-Methylcyclohexanecarbaldehyde CH3CH(OCH3)CHO a-Methoxypropionaldehyde 2-Methoxypropanal CH3CH2CH2CH2CHO Valeraldehyde Pentanal CH2=CHCHO Acrolein Prop-2-enal Phthaldehyde Benzene-1,2-dicarbaldehyde 3-Bromobenzenecarbaldehyde or m-Bromobenzaldehyde 3-Bromobenzaldehyde Ketones CH3COCH2CH2CH3 Methyl n-propyl ketone Pentan-2-one (CH3)2CHCOCH(CH3)2 Diisopropyl ketone 2,4-Dimethylpentan-3-one a-Methylcyclohexanone 2-Methylcyclohexanone (CH3)2C=CHCOCH3 Mesityl oxide 4-Methylpent-3-en-2-one Chemistry 230 Reprint 2025-26 8.1.2 Structure of The carbonyl carbon atom is sp2-hybridised and forms three sigma (s) the bonds. The fourth valence electron of carbon remains in its p-orbital Carbonyl and forms a p-bond with oxygen by overlap with p-orbital of an oxygen. Group In addition, the oxygen atom also has two non bonding electron pairs. Thus, the carbonyl carbon and the three atoms attached to it lie in the same plane and the p-electron cloud is above and below this plane. The bond angles are approximately 120° as expected of a trigonal coplanar structure (Figure 8.1). π Fig.8.1 Orbital diagram for the formation of carbonyl group The carbon-oxygen double bond is polarised due to higher electronegativity of oxygen relative to carbon. Hence, the carbonyl carbon is an electrophilic (Lewis acid), and carbonyl oxygen, a nucleophilic (Lewis base) centre. Carbonyl compounds have substantial dipole moments and are polar than ethers. The high polarity of the carbonyl group is explained on the basis of resonance involving a neutral (A) and a dipolar (B) structures as shown. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 8.1 Write the structures of the following compounds. (i) a-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec. butyl ketone (vi) 4-Fluoroacetophenone 8.28.28.28.28.2 PreparationPreparationPreparationPreparationPreparation ofofofofof AldehydesAldehydesAldehydesAldehydesAldehydes Some important methods for the preparation of aldehydes andandandandand KetonesKetonesKetonesKetonesKetones and ketones are as follows: 8.2.1 Preparation 1. By oxidation of alcohols of Aldehydes and ketones are generally prepared by oxidation of primary Aldehydes and secondary alcohols, respectively (Unit 7, Class XII). and 2. By dehydrogenation of alcohols Ketones This method is suitable for volatile alcohols and is of industrial application. In this method alcohol vapours are passed over heavy metal catalysts (Ag or Cu). Primary and secondary alcohols give aldehydes and ketones, respectively (Unit 7, Class XII). 3. From hydrocarbons (i) By ozonolysis of alkenes: As we know, ozonolysis of alkenes followed by reaction with zinc dust and water gives aldehydes, 231 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 ketones or a mixture of both depending on the substitution pattern of the alkene (Unit 9, Class XI). (ii) By hydration of alkynes: Addition of water to ethyne in the presence of H2SO4 and HgSO4 gives acetaldehyde. All other alkynes give ketones in this reaction (Unit 9, Class XI). 8.2.2 Preparation 1. From acyl chloride (acid chloride) of Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium Aldehydes on barium sulphate. This reaction is called Rosenmund reduction. 2. From nitriles and esters Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde. This reaction is called Stephen reaction. Alternatively, nitriles are selectively reduced by diisobutylaluminium hydride, (DIBAL-H) to imines followed by hydrolysis to aldehydes: Similarly, esters are also reduced to aldehydes with DIBAL-H. 3. From hydrocarbons Aromatic aldehydes (benzaldehyde and its derivatives) are prepared from aromatic hydrocarbons by the following methods: (i) By oxidation of methylbenzene Strong oxidising agents oxidise toluene and its derivatives to benzoic acids. However, it is possible to stop the oxidation at the aldehyde stage with suitable reagents that convert the methyl group to an intermediate that is difficult to oxidise further. The following methods are used for this purpose. (a) Use of chromyl chloride (CrO2Cl2): Chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde. Chemistry 232 Reprint 2025-26 This reaction is called Etard reaction. (b) Use of chromic oxide (CrO3): Toluene or substituted toluene is converted to benzylidene diacetate on treating with chromic oxide in acetic anhydride. The benzylidene diacetate can be hydrolysed to corresponding benzaldehyde with aqueous acid. (ii) By side chain chlorination followed by hydrolysis Side chain chlorination of toluene gives benzal chloride, which on hydrolysis gives benzaldehyde. This is a commercial method of manufacture of benzaldehyde. (iii) By Gatterman – Koch reaction When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives benzaldehyde or substituted benzaldehyde. This reaction is known as Gatterman-Koch reaction. 8.2.3 Preparation 1. From acyl chlorides of Ketones Treatment of acyl chlorides with dialkylcadmium, prepared by the reaction of cadmium chloride with Grignard reagent, gives ketones. 233 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 2. From nitriles Treating a nitrile with Grignard reagent followed by hydrolysis yields a ketone. 3. From benzene or substituted benzenes When benzene or substituted benzene is treated with acid chloride in the presence of anhydrous aluminium chloride, it affords the corresponding ketone. This reaction is known as Friedel-Crafts acylation reaction. ExampleExampleExampleExampleExample 8.18.18.18.18.1 Give names of the reagents to bring about the following transformations: (i) Hexan-1-ol to hexanal (ii) Cyclohexanol to cyclohexanone (iii) p-Fluorotoluene to (iv) Ethanenitrile to ethanal p-fluorobenzaldehyde (v) Allyl alcohol to propenal (vi) But-2-ene to ethanal SolutionSolutionSolutionSolutionSolution (i) C5H5NH+CrO3Cl-(PCC) (ii) Anhydrous CrO3 (iii) CrO3 in the presence (iv) (Diisobutyl)aluminium of acetic anhydride/ hydride (DIBAL-H) 1. CrO2Cl2 2. HOH (v) PCC (vi) O3/H2O-Zn dust IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 8.2 Write the structures of products of the following reactions; (C6H 5CH 2) 2 Cd + 2 CH 3 COCl (i) (ii) CH3 2 Cl2 Hg2+, H2SO 4 1.CrO (iii) H 3C C C H (iv) 2.H3O+ NO2 Chemistry 234 Reprint 2025-26 8.38.38.38.38.3 PhysicalPhysicalPhysicalPhysicalPhysical The physical properties of aldehydes and ketones are described as follows. PropertiesPropertiesPropertiesPropertiesProperties Methanal is a gas at room temperature. Ethanal is a volatile liquid. Other aldehydes and ketones are liquid or solid at room temperature. The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses. It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions. Also, their boiling points are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding. The following compounds of molecular masses 58 and 60 are ranked in order of increasing boiling points. b.p.(K) Molecular Mass n-Butane 273 58 Methoxyethane 281 60 Propanal 322 58 Acetone 329 58 Propan-1-ol 370 60 The lower members of aldehydes and ketones such as methanal, ethanal and propanone are miscible with water in all proportions, because they form hydrogen bond with water. However, the solubility of aldehydes and ketones decreases rapidly on increasing the length of alkyl chain. All aldehydes and ketones are fairly soluble in organic solvents like benzene, ether, methanol, chloroform, etc. The lower aldehydes have sharp pungent odours. As the size of the molecule increases, the odour becomes less pungent and more fragrant. In fact, many naturally occurring aldehydes and ketones are used in the blending of perfumes and flavouring agents. Arrange the following compounds in the increasing order of their ExampleExampleExampleExampleExample 8.28.28.28.28.2 boiling points: CH3CH2CH2CHO, CH3CH2CH2CH2OH, H5C2-O-C2H5, CH3CH2CH2CH3 The molecular masses of these compounds are in the range of 72 to SolutionSolutionSolutionSolutionSolution 74. Since only butan-1-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest. Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former. n-Pentane molecules have only weak van der Waals forces. Hence increasing order of boiling points of the given compounds is as follows: CH3CH2CH2CH3 < H5C2-O-C2H5 < CH3CH2CH2CHO < CH3CH2CH2CH2OH 235 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 8.3 Arrange the following compounds in increasing order of their boiling points. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3 8.48.48.48.48.4 ChemicalChemicalChemicalChemicalChemical Since aldehydes and ketones both possess the carbonyl functional group, they undergo similar chemical reactions. ReactionsReactionsReactionsReactionsReactions 1. Nucleophilic addition reactions Contrary to electrophilic addition reactions observed in alkenes, the aldehydes and ketones undergo nucleophilic addition reactions. (i) Mechanism of nucleophilic addition reactions A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp 2 hybridised orbitals of carbonyl carbon (Fig. 8.2). The hybridisation of carbon changes from sp 2 to sp 3 in this process, and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product. The net result is addition of Nu – and H + across the carbon oxygen double bond as shown in Fig. 8.2. Fig.8.2: Nucleophilic attack on carbonyl carbon (ii) Reactivity Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent. Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former. ExampleExampleExampleExampleExample 8.38.38.38.38.3 Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer. SolutionSolutionSolutionSolutionSolution The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal. Chemistry 236 Reprint 2025-26 (iii) Some important examples of nucleophilic addition and nucleophilic addition-elimination reactions: (a) Addition of hydrogen cyanide (HCN): Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins. This reaction occurs very slowly with pure HCN. Therefore, it is catalysed by a base and the generated cyanide ion (CN -) being a stronger nucleophile readily adds to carbonyl compounds to yield corresponding cyanohydrin. Cyanohydrins are useful synthetic intermediates. (b) Addition of sodium hydrogensulphite: Sodium hydrogensulphite adds to aldehydes and ketones to form the addition products. The position of the equilibrium lies largely to the right hand side for most aldehydes and to the left for most ketones due to steric reasons. The hydrogensulphite addition compound is water soluble and can be converted back to the original carbonyl compound by treating it with dilute mineral acid or alkali. Therefore, these are useful for separation and purification of aldehydes. (c) Addition of Grignard reagents: (refer Unit 7, Class XII). (d) Addition of alcohols: Aldehydes react with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to yield alkoxyalcohol intermediate, known as hemiacetals, which further react with one more molecule of alcohol to give a gem-dialkoxy compound known as acetal as shown in the reaction. Ketones react with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals. Dry hydrogen chloride protonates the oxygen of the carbonyl compounds and therefore, increases the electrophilicity of the carbonyl carbon facilitating 237 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 the nucleophilic attack of ethylene glycol. Acetals and ketals are hydrolysed with aqueous mineral acids to yield corresponding aldehydes and ketones respectively. (e) Addition of ammonia and its derivatives: Nucleophiles, such as ammonia and its derivatives H2N-Z add to the carbonyl group of aldehydes and ketones. The reaction is reversible and catalysed by acid. The equilibrium favours the product formation due to rapid dehydration of the intermediate to form >C=N-Z. Z = Alkyl, aryl, OH, NH2, C6H5NH, NHCONH2, etc. Table 8.2: Some N-Substituted Derivatives of Aldehydes and Ketones (>C=N-Z) Z Reagent name Carbonyl derivative Product name -H Ammonia Imine -R Amine Substituted imine (Schiff’s base) —OH Hydroxylamine Oxime —NH2 Hydrazine Hydrazone Phenylhydrazine Phenylhydrazone 2,4-Dinitrophenyl- 2,4 Dinitrophenyl- hydrazine hydrazone Semicarbazide Semicarbazone * 2,4-DNP-derivatives are yellow, orange or red solids, useful for characterisation of aldehydes and ketones. 2. Reduction (i) Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4) as well as by catalytic hydrogenation (Unit 7, Class XII). (ii) Reduction to hydrocarbons: The carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with zinc- amalgam and concentrated hydrochloric acid [Clemmensen Chemistry 238 Reprint 2025-26 reduction] or with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol (Wolff-Kishner reduction). 3. Oxidation Aldehydes differ from ketones in their oxidation reactions. Aldehydes Bernhard Tollens are easily oxidised to carboxylic acids on treatment with common (1841-1918) was a oxidising agents like nitric acid, potassium permanganate, potassium Professor of Chemistry dichromate, etc. Even mild oxidising agents, mainly Tollens’ reagent at the University of and Fehlings’ reagent also oxidise aldehydes. Gottingen, Germany. Ketones are generally oxidised under vigorous conditions, i.e., strong oxidising agents and at elevated temperatures. Their oxidation involves carbon-carbon bond cleavage to afford a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone. The mild oxidising agents given below are used to distinguish aldehydes from ketones: (i) Tollens’ test: On warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal. The aldehydes are oxidised to corresponding carboxylate anion. The reaction occurs in alkaline medium. (ii) Fehling’s test: Fehling reagent comprises of two solutions, Fehling solution A and Fehling solution B. Fehling solution A is aqueous copper sulphate and Fehling solution B is alkaline sodium potassium tartarate (Rochelle salt). These two solutions are mixed in equal amounts before test. On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to corresponding carboxylate anion. Aromatic aldehydes do not respond to this test. 239 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 (iii) Oxidation of methyl ketones by haloform reaction: Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform. This oxidation does not affect a carbon-carbon double bond, if present in the molecule. Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation. ExampleExampleExampleExampleExample 8.48.48.48.48.4 An organic compound (A) with molecular formula C8H8O forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens’ or Fehlings’ reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C7H6O2. Identify the compounds (A) and (B) and explain the reactions involved. SolutionSolutionSolutionSolutionSolution (A) forms 2,4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ or Fehling reagent, (A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent. This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows: Chemistry 240 Reprint 2025-26 4. Reactions due to a-hydrogen Acidity of α-hydrogens of aldehydes and ketones: The aldehydes and ketones undergo a number of reactions due to the acidic nature of α-hydrogen. The acidity of α-hydrogen atoms of carbonyl compounds is due to the strong electron withdrawing effect of the carbonyl group and resonance stabilisation of the conjugate base. (i) Aldol condensation: Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively. This is known as Aldol reaction. The name aldol is derived from the names of the two functional groups, aldehyde and alcohol, present in the products. The aldol and ketol readily lose water to give α,β-unsaturated carbonyl compounds which are aldol condensation products and the reaction is called Aldol condensation. Though ketones give ketols (compounds containing a keto and alcohol groups), the general name aldol condensation still applies to the reactions of ketones due to their similarity with aldehydes. 241 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 (ii) Cross aldol condensation: When aldol condensation is carried out between two different aldehydes and / or ketones, it is called cross aldol condensation. If both of them contain a-hydrogen atoms, it gives a mixture of four products. This is illustrated below by aldol reaction of a mixture of ethanal and propanal. Ketones can also be used as one component in the cross aldol reactions. 5. Other reactions (i) Cannizzaro reaction: Aldehydes which do not have an a-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on heating with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt. D D Chemistry 242 Reprint 2025-26 (ii) Electrophilic substitution reaction: Aromatic aldehydes and ketones undergo electrophilic substitution at the ring in which the carbonyl group acts as a deactivating and meta-directing group. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 8.4 Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Ethanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. Hint: Consider steric effect and electronic effect. 8.5 Predict the products of the following reactions: (i) (ii) (iii) (iv) 8.58.58.58.58.5 UsesUsesUsesUsesUses ofofofofof In chemical industry aldehydes and ketones are used as solvents, AldehydesAldehydesAldehydesAldehydesAldehydes starting materials and reagents for the synthesis of other products. Formaldehyde is well known as formalin (40%) solution used to preserve andandandandand KetonesKetonesKetonesKetonesKetones biological specimens and to prepare bakelite (a phenol-formaldehyde resin), urea-formaldehyde glues and other polymeric products. Acetaldehyde is used primarily as a starting material in the manufacture of acetic acid, ethyl acetate, vinyl acetate, polymers and drugs. Benzaldehyde is used in perfumery and in dye industries. Acetone and ethyl methyl ketone are common industrial solvents. Many aldehydes and ketones, e.g., butyraldehyde, vanillin, acetophenone, camphor, etc. are well known for their odours and flavours. 243 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 Carboxylic Acids Carbon compounds containing a carboxyl functional group, –COOH are called carboxylic acids. The carboxyl group, consists of a carbonyl group attached to a hydroxyl group, hence its name carboxyl. Carboxylic acids may be aliphatic (RCOOH) or aromatic (ArCOOH) depending on the group, alkyl or aryl, attached to carboxylic carbon. Large number of carboxylic acids are found in nature. Some higher members of aliphatic carboxylic acids (C12 – C18) known as fatty acids, occur in natural fats as esters of glycerol. Carboxylic acids serve as starting material for several other important organic compounds such as anhydrides, esters, acid chlorides, amides, etc. 8.68.68.68.68.6 NomenclatureNomenclatureNomenclatureNomenclatureNomenclature andandandandand StructureStructureStructureStructureStructure ofofofofof CarboxylCarboxylCarboxylCarboxylCarboxyl GroupGroupGroupGroupGroup 8.6.1 Since carboxylic acids are amongst the earliest organic compounds to Nomenclature be isolated from nature, a large number of them are known by their common names. The common names end with the suffix –ic acid and have been derived from Latin or Greek names of their natural sources. For example, formic acid (HCOOH) was first obtained from red ants (Latin: formica means ant), acetic acid (CH3COOH) from vinegar (Latin: acetum, means vinegar), butyric acid (CH3CH2CH2COOH) from rancid butter (Latin: butyrum, means butter). In the IUPAC system, aliphatic carboxylic acids are named by replacing the ending –e in the name of the corresponding alkane with – oic acid. In numbering the carbon chain, the carboxylic carbon is numbered one. For naming compounds containing more than one carboxyl group, the alkyl chain leaving carboxyl groups is numbered and the number of carboxyl groups is indicated by adding the multiplicative prefix, dicarboxylic acid, tricarboxylic acid, etc. to the name of parent alkyl chain. The position of –COOH groups are indicated by the arabic numeral before the multiplicative prefix. Some of the carboxylic acids along with their common and IUPAC names are listed in Table 8.3. Table 8.3 Names and Structures of Some Carboxylic Acids Structure Common name IUPAC name HCOOH Formic acid Methanoic acid CH3COOH Acetic acid Ethanoic acid CH3CH2COOH Propionic acid Propanoic acid CH3CH2CH2COOH Butyric acid Butanoic acid (CH3)2CHCOOH Isobutyric acid 2-Methylpropanoic acid HOOC-COOH Oxalic acid Ethanedioic acid HOOC -CH2-COOH Malonic acid Propanedioic acid HOOC -(CH2)2-COOH Succinic acid Butanedioic acid HOOC -(CH2)3-COOH Glutaric acid Pentanedioic acid HOOC -(CH2)4-COOH Adipic acid Hexanedioic acid HOOC -CH2-CH(COOH)-CH2-COOH Tricarballylic acid Propane-1, 2, 3- or carballylic acid tricarboxylic acid Chemistry 244 Reprint 2025-26 Benzoic acid Benzenecarboxylic acid (Benzoic acid) Phenylacetic acid 2-Phenylethanoic acid Phthalic acid Benzene-1, 2-dicarboxylic acid 8.6.2 Structure In carboxylic acids, the bonds to the carboxyl carbon lie in one plane of Carboxyl and are separated by about 120°. The carboxylic carbon is less Group electrophilic than carbonyl carbon because of the possible resonance structure shown below: + IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 8.6 Give the IUPAC names of the following compounds: (i) Ph CH2CH2COOH (ii) (CH3)2C=CHCOOH CH3 (iii) COOH (iv) 8.78.78.78.78.7 MethodsMethodsMethodsMethodsMethods ofofofofof Some important methods of preparation of carboxylic acids are as follows. PreparationPreparationPreparationPreparationPreparation 1. From primary alcohols and aldehydes ofofofofof CarboxylicCarboxylicCarboxylicCarboxylicCarboxylic Primary alcohols are readily oxidised to carboxylic acids with common AcidsAcidsAcidsAcidsAcids oxidising agents such as potassium permanganate (KMnO4) in neutral, acidic or alkaline media or by potassium dichromate (K2Cr2O7) and chromium trioxide (CrO3) in acidic media (Jones reagent). Jones reagent 245 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 Carboxylic acids are also prepared from aldehydes by the use of mild oxidising agents (Section 8.4). 2. From alkylbenzenes Aromatic carboxylic acids can be prepared by vigorous oxidation of alkyl benzenes with chromic acid or acidic or alkaline potassium permanganate. The entire side chain is oxidised to the carboxyl group irrespective of length of the side chain. Primary and secondary alkyl groups are oxidised in this manner while tertiary group is not affected. Suitably substituted alkenes are also oxidised to carboxylic acids with these oxidising reagents. 3. From nitriles and amides Nitriles are hydrolysed to amides and then to acids in the presence of  H+ or OH as catalyst. Mild reaction conditions are used to stop the reaction at the amide stage. 4. From Grignard reagents Grignard reagents react with carbon dioxide (dry ice) to form salts of carboxylic acids which in turn give corresponding carboxylic acids after acidification with mineral acid. As we know, the Grignard reagents and nitriles can be prepared from alkyl halides (refer Unit 6, Class XII). The above methods Chemistry 246 Reprint 2025-26 (3 and 4) are useful for converting alkyl halides into corresponding carboxylic acids having one carbon atom more than that present in alkyl halides (ascending the series). 5. From acyl halides and anhydrides Acid chlorides when hydrolysed with water give carboxylic acids or more readily hydrolysed with aqueous base to give carboxylate ions which on acidification provide corresponding carboxylic acids. Anhydrides on the other hand are hydrolysed to corresponding acid(s) with water. 6. From esters Acidic hydrolysis of esters gives directly carboxylic acids while basic hydrolysis gives carboxylates, which on acidification give corresponding carboxylic acids. Write chemical reactions to affect the following transformations: ExampleExampleExampleExampleExample 8.58.58.58.58.5 (i) Butan-1-ol to butanoic acid (ii) Benzyl alcohol to phenylethanoic acid (iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid (iv) 4-Methylacetophenone to benzene-1,4-dicarboxylic acid (v) Cyclohexene to hexane-1,6-dioic acid (vi) Butanal to butanoic acid. 247 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 SolutionSolutionSolutionSolutionSolution (i) (ii) (iii) (iv) (v) (vi) IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 8.7 Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene) Chemistry 248 Reprint 2025-26 8.88.88.88.88.8 PhysicalPhysicalPhysicalPhysicalPhysical Aliphatic carboxylic acids upto nine carbon atoms are colourless PropertiesPropertiesPropertiesPropertiesProperties liquids at room temperature with unpleasant odours. The higher acids are wax like solids and are practically odourless due to their low volatility. Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding. The hydrogen bonds are not broken completely even in the vapour phase. In fact, In vapour state or in most carboxylic acids exist as dimer in the vapour phase aprotic solvent or in the aprotic solvents. Simple aliphatic carboxylic acids having upto four carbon atoms are miscible in water due to the formation of hydrogen bonds with water. The solubility decreases with increasing number of carbon atoms. Higher carboxylic acids are practically insoluble in water due to the increased hydrophobic interaction of hydrocarbon part. Benzoic acid, the simplest aromatic carboxylic acid is nearly insoluble in cold water. Carboxylic acids are Hydrogen bonding of also soluble in less polar organic solvents like benzene, RCOOH with H2O ether, alcohol, chloroform, etc. 8.98.98.98.98.9 ChemicalChemicalChemicalChemicalChemical ReactionsReactionsReactionsReactionsReactions The reaction of carboxylic acids are classified as follows: 8.9.1 Reactions Acidity Involving Reactions with metals and alkalies Cleavage of The carboxylic acids like alcohols evolve hydrogen with electropositive O–H Bond metals and form salts with alkalies similar to phenols. However, unlike phenols they react with weaker bases such as carbonates and hydrogencarbonates to evolve carbon dioxide. This reaction is used to detect the presence of carboxyl group in an organic compound. Carboxylic acids dissociate in water to give resonance stabilised carboxylate anions and hydronium ion. 249 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 For the above reaction: where Keq, is equilibrium constant and Ka is the acid dissociation constant. For convenience, the strength of an acid is generally indicated by its pKa value rather than its Ka value. pKa = – log Ka The pKa of hydrochloric acid is –7.0, where as pKa of trifluoroacetic acid (the strongest carboxylic acid), benzoic acid and acetic acid are 0.23, 4.19 and 4.76, respectively. Smaller the pKa, the stronger the acid ( the better it is as a proton donor). Strong acids have pKa values < 1, the acids with pKa values between 1 and 5 are considered to be moderately strong acids, weak acids have pKa values between 5 and 15, and extremely weak acids have pKa values >15. Carboxylic acids are weaker than mineral acids, but they are stronger acids than alcohols and many simple phenols (pKa is ~16 for ethanol and 10 for phenol). In fact, carboxylic acids are amongst the most acidic organic compounds you have studied so far. You already know why phenols are more acidic than alcohols. The higher acidity of carboxylic acids as compared to phenols can be understood similarly. The conjugate base of carboxylic acid, a carboxylate ion, is stabilised by two equivalent resonance structures in which the negative charge is at the more electronegative oxygen atom. The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom. Therefore, resonance in phenoxide ion is not as important as it is in carboxylate ion. Further, the negative charge is delocalised over two electronegative oxygen atoms in carboxylate ion whereas it is less effectively delocalised over one oxygen atom and less electronegative carbon atoms in phenoxide ion (Unit 7, Class XII). Thus, the carboxylate ion is more stabilised than phenoxide ion, so carboxylic acids are more acidic than phenols. Effect of substituents on the acidity of carboxylic acids: Substituents may affect the stability of the conjugate base and thus, also affect the acidity of the carboxylic acids. Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive and/or resonance effects. Conversely, electron donating groups decrease the acidity by destabilising the conjugate base. Electron withdrawing group (EWG) Electron donating group (EDG) stabilises the carboxylate anion destabilises the carboxylate and strengthens the acid anion and weakens the acid Chemistry 250 Reprint 2025-26 The effect of the following groups in increasing acidity order is Ph < I < Br < Cl < F < CN < NO2 < CF3 Thus, the following acids are arranged in order of increasing acidity (based on pKa values): CF3COOH > CCl3COOH > CHCl2COOH > NO2CH2COOH > NC-CH2COOH > FCH2COOH > ClCH2COOH > BrCH2COOH > HCOOH > ClCH2CH2COOH > (continue) C6H5COOH > C6H5CH2COOH > CH3COOH > CH3CH2COOH (continue ) Direct attachment of groups such as phenyl or vinyl to the carboxylic acid, increases the acidity of corresponding carboxylic acid, contrary to the decrease expected due to resonance effect shown below: This is because of greater electronegativity of sp2 hybridised carbon to which carboxyl carbon is attached. The presence of electron withdrawing group on the phenyl of aromatic carboxylic acid increases their acidity while electron donating groups decrease their acidity. COOH COOH COOH NO2 OCH3 4-Methoxy Benzoic acid 4-Nitrobenzoic acid benzoic acid (p Ka = 4.46) (p Ka = 4.19) (p Ka = 3.41) 8.9.2 Reactions 1. Formation of anhydride Involving Carboxylic acids on heating with mineral acids such as H2SO4 or with Cleavage of P2O5 give corresponding anhydride. C–OH Bond 2. Esterification Carboxylic acids are esterified with alcohols or phenols in the presence of a mineral acid such as concentrated H2SO4 or HCl gas as a catalyst. 251 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 Mechanism of esterification of carboxylic acids: The esterification of carboxylic acids with alcohols is a kind of nucleophilic acyl substitution. Protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol. Proton transfer in the tetrahedral intermediate converts the hydroxyl group into – +OH2 group, which, being a better leaving group, is eliminated as neutral water molecule. The protonated ester so formed finally loses a proton to give the ester. 3. Reactions with PCl5, PCl3 and SOCl2 The hydroxyl group of carboxylic acids, behaves like that of alcohols and is easily replaced by chlorine atom on treating with PCl5, PCl3 or SOCl2. Thionyl chloride (SOCl2) is preferred because the other two products are gaseous and escape the reaction mixture making the purification of the products easier. 4. Reaction with ammonia Carboxylic acids react with ammonia to give ammonium salt which on further heating at high temperature give amides. For example: Chemistry 252 Reprint 2025-26 8.9.3 Reactions 1. Reduction Involving Carboxylic acids are reduced to primary alcohols by lithium –COOH aluminium hydride or better with diborane. Diborane does not easily Group reduce functional groups such as ester, nitro, halo, etc. Sodium borohydride does not reduce the carboxyl group. 2. Decarboxylation Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaO in the ratio of 3 : 1). The reaction is known as decarboxylation. Alkali metal salts of carboxylic acids also undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid. The reaction is known as Kolbe electrolysis (Unit 9, Class XI). 8.9.4 Substitution 1. Halogenation Reactions in the Carboxylic acids having an α-hydrogen are halogenated at the Hydrocarbon Part α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids. The reaction is known as Hell-Volhard-Zelinsky reaction. 253 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 2. Ring substitution Aromatic carboxylic acids undergo electrophilic substitution reactions in which the carboxyl group acts as a deactivating and meta-directing group. They however, do not undergo Friedel-Crafts reaction (because the carboxyl group is deactivating and the catalyst aluminium chloride (Lewis acid) gets bonded to the carboxyl group). IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 8.8 Which acid of each pair shown here would you expect to be stronger? (i) CH3CO2H or CH2FCO2H (ii) CH2FCO2H or CH2ClCO2H (iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H (iv) 8.108.108.108.108.10 UsesUsesUsesUsesUses ofofofofof Methanoic acid is used in rubber, textile, dyeing, leather and electroplating CarboxylicCarboxylicCarboxylicCarboxylicCarboxylic industries. Ethanoic acid is used as solvent and as vinegar in food industry. Hexanedioic acid is used in the manufacture of nylon-6, 6. Esters of benzoic AcidsAcidsAcidsAcidsAcids acid are used in perfumery. Sodium benzoate is used as a food preservative. Higher fatty acids are used for the manufacture of soaps and detergents. SummarySummarySummarySummarySummary Aldehydes, ketones and carboxylic acids are some of the important classes of organic compounds containing carbonyl group. These are highly polar molecules. Therefore, they boil at higher temperatures than the hydrocarbons and weakly polar compounds such as ethers of comparable molecular masses. The lower members are more soluble in water because they form hydrogen bonds with water. The higher members, because of large size of hydrophobic chain of carbon atoms, are insoluble in water but soluble in common organic solvents. Aldehydes are prepared by dehydrogenation or controlled oxidation of primary alcohols and controlled or selective reduction of acyl halides. Aromatic aldehydes may also be prepared by oxidation of (i) methylbenzene with chromyl chloride or CrO3 in the presence of acetic anhydride, (ii) formylation of arenes with carbon monoxide and hydrochloric acid in the presence of anhydrous aluminium chloride, and (iii) cuprous chloride or by hydrolysis of benzal chloride. Ketones are prepared by oxidation of secondary alcohols and hydration of alkynes. Ketones are also prepared by reaction of acyl chloride with dialkylcadmium. A good method for the preparation of aromatic ketones is the Friedel-Crafts acylation of aromatic hydrocarbons with acyl chlorides or anhydrides. Both aldehydes and ketones can be prepared by ozonolysis of alkenes. Aldehydes and ketones undergo nucleophilic addition reactions onto the carbonyl group with a number of nucleophiles such as, HCN, NaHSO3, alcohols (or diols), Chemistry 254 Reprint 2025-26 ammonia derivatives, and Grignard reagents. The a-hydrogens in aldehydes and ketones are acidic. Therefore, aldehydes and ketones having at least one a-hydrogen, undergo Aldol condensation in the presence of a base to give a-hydroxyaldehydes (aldol) and a-hydroxyketones(ketol), respectively. Aldehydes having no a-hydrogen undergo Cannizzaro reaction in the presence of concentrated alkali. Aldehydes and ketones are reduced to alcohols with NaBH4, LiAlH4, or by catalytic hydrogenation. The carbonyl group of aldehydes and ketones can be reduced to a methylene group by Clemmensen reduction or Wolff-Kishner reduction. Aldehydes are easily oxidised to carboxylic acids by mild oxidising reagents such as Tollens’ reagent and Fehling’s reagent. These oxidation reactions are used to distinguish aldehydes from ketones. Carboxylic acids are prepared by the oxidation of primary alcohols, aldehydes and alkenes by hydrolysis of nitriles, and by treatment of Grignard reagents with carbon dioxide. Aromatic carboxylic acids are also prepared by side-chain oxidation of alkylbenzenes. Carboxylic acids are considerably more acidic than alcohols and most of simple phenols. Carboxylic acids are reduced to primary alcohols with LiAlH4, or better with diborane in ether solution and also undergo a-halogenation with Cl2 and Br2 in the presence of red phosphorus (Hell-Volhard Zelinsky reaction). Methanal, ethanal, propanone, benzaldehyde, formic acid, acetic acid and benzoic acid are highly useful compounds in industry. Exercises

7.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane. 223 Alcohols, Phenols and Ethers Reprint 2025-26

7.18 Explain the following with an example. (i) Kolbe’s reaction. (ii) Reimer-Tiemann reaction. (iii) Williamson ether synthesis. (iv) Unsymmetrical ether.

7.10 HBr C2 H5 OH C2 H5 Br CH3 – CH2 – CH – CH – ONa + C 2H5 Br CH3 – CH2 – CH – CH – OC2 H5 CH3 CH3 CH3 CH3 2-Ethoxy-3-methylpentane 225 Alcohols, Phenols and Ethers Reprint 2025-26

7.9 Write the equations involved in the following reactions: (i) Reimer - Tiemann reaction (ii) Kolbe’s reaction 213 Alcohols, Phenols and Ethers Reprint 2025-26 7.57.57.57.57.5 SomeSomeSomeSomeSome Methanol and ethanol are among the two commercially important alcohols. CommerciallyCommerciallyCommerciallyCommerciallyCommercially ImportantImportantImportantImportantImportant 1. Methanol AlcoholsAlcoholsAlcoholsAlcoholsAlcohols Methanol, CH3OH, also known as ‘wood spirit’, was produced by destructive distillation of wood. Today, most of the methanol is produced by catalytic hydrogenation of carbon monoxide at high pressure and temperature and in the presence of ZnO – Cr2O3 catalyst. Methanol is a colourless liquid and boils at 337 K. It is highly poisonous in nature. Ingestion of even small quantities of methanol can cause blindness and large quantities causes even death. Methanol is used as a solvent in paints, varnishes and chiefly for making formaldehyde. 2. Ethanol Ethanol, C2H5OH, is obtained commercially by fermentation, the oldest method is from sugars. The sugar in molasses, sugarcane or fruits such as grapes is converted to glucose and fructose, (both of which have the formula C6H12O6), in the presence of an enzyme, invertase. Glucose and fructose undergo fermentation in the presence of another enzyme, zymase, which is found in yeast. In wine making, grapes are the source of sugars and yeast. As Ingestion of ethanol acts grapes ripen, the quantity of sugar increases and yeast grows on the on the central nervous outer skin. When grapes are crushed, sugar and the enzyme come in system. In moderate contact and fermentation starts. Fermentation takes place in amounts, it affects anaerobic conditions i.e. in absence of air. Carbon dioxide is released judgment and lowers during fermentation. inhibitions. Higher concentrations cause The action of zymase is inhibited once the percentage of alcohol nausea and loss of formed exceeds 14 percent. If air gets into fermentation mixture, the consciousness. Even at oxygen of air oxidises ethanol to ethanoic acid which in turn destroys higher concentrations, the taste of alcoholic drinks. it interferes with Ethanol is a colourless liquid with boiling point 351 K. It is used spontaneous respiration as a solvent in paint industry and in the preparation of a number of and can be fatal. carbon compounds. The commercial alcohol is made unfit for drinking by mixing in it some copper sulphate (to give it a colour) and pyridine (a foul smelling liquid). It is known as denaturation of alcohol. Nowadays, large quantities of ethanol are obtained by hydration of ethene (Section 7.4). Chemistry 214 Reprint 2025-26 7.67.67.67.67.6 EthersEthersEthersEthersEthers 7.6.1 Preparation 1. By dehydration of alcohols of Ethers Alcohols undergo dehydration in the presence of protic acids (H2SO4, H3PO4). The formation of the reaction product, alkene or ether depends on the reaction conditions. For example, ethanol is dehydrated to ethene in the presence of sulphuric acid at 443 K. At 413 K, ethoxyethane is the main product. Diethyl ether has been used widely as an inhalation anaesthetic. But due to its slow effect and an unpleasant recovery The formation of ether is a nucleophilic bimolecular reaction (SN2) period, it has been involving the attack of alcohol molecule on a protonated alcohol, as replaced, as an indicated below: anaesthetic, by other compounds. Acidic dehydration of alcohols, to give an alkene is also associated with substitution reaction to give an ether. The method is suitable for the preparation of ethers having primary alkyl groups only. The alkyl group should be unhindered and the temperature be kept low. Otherwise the reaction favours the formation of alkene. The reaction follows SN1 pathway when the alcohol is secondary or tertiary about which you will learn in higher classes. However, the dehydration of secondary and tertiary alcohols to give corresponding ethers is unsuccessful as elimination competes over substitution and as a consequence, alkenes are easily formed. Can you explain why is bimolecular dehydration not appropriate for the preparation of ethyl methyl ether? 2. Williamson synthesis It is an important laboratory method for the preparation of Alexander William symmetrical and unsymmetrical ethers. In this method, an alkyl Williamson (1824–1904) halide is allowed to react with sodium alkoxide.was born in London of – +Scottish parents. In R–X + R’–O Na R–O–R’ + Na X 1849, he became Professor of Chemistry Ethers containing substituted alkyl groups (secondary or tertiary) at University College, may also be prepared by this method. The reaction involves SN2 attack London. of an alkoxide ion on primary alkyl halide. 215 Alcohols, Phenols and Ethers Reprint 2025-26 O Na + CH 3–Br Better results are obtained if the alkyl halide is primary. In case of secondary and tertiary alkyl halides, elimination competes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed. For example, the reaction of CH3ONa with (CH3)3C–Br gives exclusively 2-methylpropene. It is because alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to elimination reactions. ExampleExampleExampleExampleExample 7.67.67.67.67.6 The following is not an appropriate reaction for the preparation of t-butyl ethyl ether. (i) What would be the major product of this reaction ? (ii) Write a suitable reaction for the preparation of t-butylethyl ether. SolutionSolutionSolutionSolutionSolution (i) The major product of the given reaction is 2-methylprop-1-ene. It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution. (ii) Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety. Chemistry 216 Reprint 2025-26 7.6.2 Physical The C-O bonds in ethers are polar and thus, ethers have a net dipole Properties moment. The weak polarity of ethers do not appreciably affect their boiling points which are comparable to those of the alkanes of comparable molecular masses but are much lower than the boiling points of alcohols as shown in the following cases: Formula CH3(CH2)3CH3 C2H5-O-C2H5 CH3(CH2)3-OH n-Pentane Ethoxyethane Butan-1-ol b.p./K 309.1 307.6 390 The large difference in boiling points of alcohols and ethers is due to the presence of hydrogen bonding in alcohols. The miscibility of ethers with water resembles those of alcohols of the same molecular mass. Both ethoxyethane and butan-1-ol are miscible to almost the same extent i.e., 7.5 and 9 g per 100 mL water, respectively while pentane is essentially immiscible with water. Can you explain this observation ? This is due to the fact that just like alcohols, oxygen of ether can also form hydrogen bonds with water molecule as shown: 7.6.3 Chemical 1. Cleavage of C–O bond in ethers Reactions Ethers are the least reactive of the functional groups. The cleavage of C-O bond in ethers takes place under drastic conditions with excess of hydrogen halides. The reaction of dialkyl ether gives two alkyl halide molecules. Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to the more stable aryl-oxygen bond. The reaction yields phenol and alkyl halide. Ethers with two different alkyl groups are also cleaved in the same manner. The order of reactivity of hydrogen halides is as follows: HI > HBr > HCl. The cleavage of ethers takes place with concentrated HI or HBr at high temperature. 217 Alcohols, Phenols and Ethers Reprint 2025-26 MechanismMechanismMechanismMechanismMechanism The reaction of an ether with concentrated HI starts with protonation of ether molecule. Step 1: The reaction takes place with HBr or HI because these reagents are sufficiently acidic. Step 2: Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN2 mechanism. Thus, in the cleavage of mixed ethers with two different alkyl groups, the alcohol and alkyl iodide formed, depend on the nature of alkyl groups. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide (SN2 reaction). When HI is in excess and the reaction is carried out at high temperature, ethanol reacts with another molecule of HI and is converted to ethyl iodide. Step 3: However, when one of the alkyl group is a tertiary group, the halide formed is a tertiary halide. CH3 CH3 CH3 C O CH3 +HI CH3OH +CH 3 C I CH3 CH3 It is because in step 2 of the reaction, the departure of leaving group (HO–CH3) creates a more stable carbocation [(CH3)3C+], and the reaction follows SN1 mechanism. In case of anisole, methylphenyl CH3 CH3 + slow + oxonium ion, is CH3 C O CH3 CH3 C + CH3 OH H CH3 CH3 formed by protonation of ether. The bond between O–CH3 is weaker CH3 CH3 than the bond between O–C6H5 + – fast CH3 C + I CH3 C I because the carbon of phenyl group is sp2 hybridised and there CH3 CH3 is a partial double bond character. Chemistry 218 Reprint 2025-26 Therefore the attack by I– ion breaks O–CH3 bond to form CH3I. Phenols do not react further to give halides because the sp 2 hybridised carbon of phenol cannot undergo nucleophilic substitution reaction needed for conversion to the halide. Give the major products that are formed by heating each of the following ExampleExampleExampleExampleExample 7.77.77.77.77.7 ethers with HI. (i) (ii) (iii) (i) (ii) SolutionSolutionSolutionSolutionSolution (iii) 2. Electrophilic substitution The alkoxy group (-OR) is ortho, para directing and activates the aromatic ring towards electrophilic substitution in the same way as in phenol. (i) Halogenation: Phenylalkyl ethers undergo usual halogenation in the benzene ring, e.g., anisole undergoes bromination with bromine in ethanoic acid even in the absence of iron (III) bromide catalyst. It is due to the activation of benzene ring by the methoxy group. Para isomer is obtained in 90% yield. 219 Alcohols, Phenols and Ethers Reprint 2025-26 (ii) Friedel-Crafts reaction: Anisole undergoes Friedel-Crafts reaction, i.e., the alkyl and acyl groups are introduced at ortho and para positions by reaction with alkyl halide and acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst. (iii) Nitration: Anisole reacts with a mixture of concentrated sulphuric and nitric acids to yield a mixture of ortho and para nitroanisole. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 7.10 Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol. 7.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? (i) (ii) Chemistry 220 Reprint 2025-26

9.9 Give the structures of A, B and C in the following reactions: B NaOH  Br 2 C (i) CH 3 CH 2 I NaCN A Partial OHhydrolysis CuCN H 2 O/H  NH 3 (ii) C 6 H 5 N 2 Cl  A  B  C 4 B HNO 2 C (iii) CH 3 CH 2 Br KCN A LiAlH 0  C  2  2 O/H HCl B H C  (iv) C6 H 5 NO 2 Fe/HCl A NaNO273 K 3 A NaOBr B NaNO 2 /HCl C (v) CH 3 COOH NH 2 B C 6 H 5 OH C (vi) C6 H 5 NO 2 Fe/HCl A HNO273K 279 Amines Reprint 2025-26

9.4 Alkynes 1 2 3 4 5 Like alkenes, alkynes are also unsaturated II. H3C–C≡ C– CH2– CH3 Pent–2-yne hydrocarbons. They contain at least one triple 4 3 2 1 bond between two carbon atoms. The number III. H3C–CH–C≡ CH 3-Methyl but–1-yne |of hydrogen atoms is still less in alkynes as CH3compared to alkenes or alkanes. Their general Structures I and II are position isomers formula is CnH2n–2. and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 9.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen corresponding to the 5 th member of gas. Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it all the isomers. What type of isomerism is interesting to study this class of organic is exhibited by different pairs of isomers? compounds. Solution 9.4.1 Nomenclature and Isomerism th 5 member of alkyne has the molecular In common system, alkynes are named as formula C6H10. The possible isomers are: derivatives of acetylene. In IUPAC system, they Table 9.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne Reprint 2025-26 Hydrocarbons 315 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 9.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C different pairs. π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol -1) is more than those 9.4.2 Structure of Triple Bond of C=C bond (bond enthalpy 681 kJ mol –1) Ethyne is the simplest molecule of alkyne and C–C bond (bond enthalpy 348 kJ mol–1). series. Structure of ethyne is shown in The C≡C bond length is shorter (120 pm) Fig. 9.6. than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon Each carbon atom of ethyne has two sp atoms is cylindrically symmetrical about thehybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 9.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating the 1s orbital of each of the two hydrogen calcium carbide with water. Calcium atoms forming two C-H sigma bonds. carbide is prepared by heating quick lime H-C-C bond angle is of 180°. Each carbon with coke. Quick lime can be obtained byhas two unhybridised p orbitals which are heating limestone as shown in the followingperpendicular to each other as well as to the reactions:plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆ CaO + O2 (9.55) Reprint 2025-26 316 chemistry CaO + 3C CaC2 + CO (9.56) the sp hybridised carbon2 atoms whereas they are attached to sp hybridised carbon Calcium 3 atoms in ethene and sp hybridised carbons carbide in ethane. Due to the maximum percentage of CaC2 + 2H2O Ca(OH)2 + C2H2 (9.57) s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have2. From vicinal dihalides : Vicinal dihalides highest electronegativity; hence, these attract on treatment with alcoholic potassium the shared electron pair of the C-H bond of hydroxide undergo dehydrohalogenation. ethyne to a greater extent than that of the One molecule of hydrogen halide is 2 sp hybridised orbitals of carbon in ethene eliminated to form alkenyl halide which 3 and the sp hybridised orbital of carbon in on treatment with sodamide gives alkyne. ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. HC ≡ CH + Na → HC ≡ C–Na++ 1/2 H2 9.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same (9.59) trend of alkenes and alkanes. First three HC ≡ C– Na + Na → Na+ C–Na+ ≡ C–Na++ 1/2 H2members are gases, the next eight are liquids and the higher ones are solids. All alkynes Disodium ethynide are colourless. Ethyene has characteristic (9.60)odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter CH3 – C ≡ C – H + Na+ NH–2 than water and immiscible with water but ↓ soluble in organic solvents like ethers, carbon CH3 – C ≡ C– Na+ + NH3 tetrachloride and benzene. Their melting Sodium propynide (9.61) point, boiling point and density increase with These reactions are not shown by alkenesincrease in molar mass. and alkanes, hence used for distinction Chemical properties between alkynes, alkenes and alkanes. What Alkynes show acidic nature, addition reactions about the above reactions with but-1-yne and and polymerisation reactions as follows : but-2-yne ? Alkanes, alkenes and alkynes A. Acidic character of alkyne: Sodium follow the following trend in their acidic metal and sodamide (NaNH2) are strong behaviour : bases. They react with ethyne to form sodium i) CH ≡ CH > H2C – CH2 > CH3 –CH3acetylide with the liberation of dihydrogen gas. These reactions have not been observed ii) HC ≡ CH > CH3 –C≡ CH >> CH3 –C≡C–CH3in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison B. Addition reactions: Alkynes contain a to ethene and ethane. Why is it so ? Has triple bond, so they add up, two molecules of it something to do with their structures dihydrogen, halogen, hydrogen halides etc. and the hybridisation ? You have read that Formation of the addition product takes place hydrogen atoms in ethyne are attached to according to the following steps. Reprint 2025-26 Hydrocarbons 317 The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions (9.66)are given below: (i) Addition of dihydrogen (iv) Addition of water Pt/Pd/Ni H2 Like alkanes and alkenes, alkynes are alsoHC≡CH+H2 [H2C = CH2] CH3–CH3 immiscible and do not react with water. (9.62) However, one molecule of water adds to alkynes on warming with mercuric sulphate CH3–C≡CH + H2 Pt/Pd/Ni [CH3–CH=CH2] and dilute sulphuric acid at 333 K to form Propyne Propene carbonyl compounds. ↓H2 CH3–CH2–CH3 Propane (9.63) (ii) Addition of halogens (9.67) (9.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (iii) Addition of hydrogen halides (9.68) Two molecules of hydrogen halides (HCl, HBr, (v) Polymerisation HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H–C≡C–H+H–Br [CH2 = CH–Br]→ CHBr2 polyethyne which is a high molecular Bromoethene weight polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —(CH = CH – CH = CH)n— Under special (9.65) conditions, this polymer conducts electricity. Reprint 2025-26 318 chemistry Thin film of polyacetylene can be used as but in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the is retained. However, there are examples of metal conductors. aromatic hydrocarbons which do not contain a (b) Cyclic polymerisation: Ethyne on benzene ring but instead contain other highly unsaturated ring. Aromatic compoundspassing through red hot iron tube at 873K containing benzene ring are known asundergoes cyclic polymerization. Three benzenoids and those not containing amolecules polymerise to form benzene, which benzene ring are known as non-benzenoids.is the starting molecule for the preparation of Some examples of arenes are givenderivatives of benzene, dyes, drugs and large below:number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (9.69) Biphenyl Problem 9.14 How will you convert ethanoic acid into 9.5.1 Nomenclature and Isomerism benzene? The nomenclature and isomerism of aromatic Solution hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:

9.5 How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid? 9.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

9.7 Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.

9.8 Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene (v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-chloroaniline (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix) Aniline to benzyl alcohol.

9.10 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C. 9.11 Complete the following reactions: (i) C6 H 5 NH 2  CHCl 3  alc.KOH  (ii) C 6 H 5 N 2 Cl  H 3 PO 2  H 2 O  (iii) C6 H 5 NH 2  H 2 SO 4  conc.  (iv) C6 H 5 N 2 Cl  C 2 H 5 OH  (v) C6 H5 NH 2  Br2  aq   O  (vi) C6 H 5 NH 2   CH 3 CO  2 4  (vii) C 6 H5 N 2 Cl ii NaNOi HBF2 /Cu, 9.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis? 9.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid. 9.14 Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines? Answers to Some Intext Questions 9.4 (i) C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH (ii) C6H5NH2 < C2H5NH2. < (C2H5)3N < (C2H5)2NH (iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH Chemistry 280 Reprint 2025-26 UnitUnitUnitUnit Unit1010 BiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesBiomoleculesObjectives After studying this Unit, you will be able to “It is the harmonious and synchronous progress of chemical • explain the characteristics of reactions in body which leads to life”. biomolecules like carbohydrates, proteins and nucleic acids and hormones; A living system grows, sustains and reproduces itself. • classify carbohydrates, proteins, The most amazing thing about a living system is that it nucleic acids and vitamins on the basis of their structures; is composed of non-living atoms and molecules. The • explain the difference between pursuit of knowledge of what goes on chemically within DNA and RNA; a living system falls in the domain of biochemistry. Living • describe the role of biomolecules systems are made up of various complex biomolecules in biosystem. like carbohydrates, proteins, nucleic acids, lipids, etc. Proteins and carbohydrates are essential constituents of our food. These biomolecules interact with each other and constitute the molecular logic of life processes. In addition, some simple molecules like vitamins and mineral salts also play an important role in the functions of organisms. Structures and functions of some of these biomolecules are discussed in this Unit. 10.110.110.110.110.1 CarbohydratesCarbohydratesCarbohydratesCarbohydratesCarbohydrates Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds. Some common examples of carbohydrates are cane sugar, glucose, starch, etc. Most of them have a general formula, Cx(H2O)y, and were considered as hydrates of carbon from where the name carbohydrate was derived. For example, the molecular formula of glucose (C6H12O6) fits into this general formula, C6(H2O)6. But all the compounds which fit into this formula may not be classified as carbohydrates. For example acetic acid (CH3COOH) fits into this general formula, C2(H2O)2 but is not a carbohydrate. Similarly, rhamnose, C6H12O5 is a carbohydrate but does not fit in this definition. A large number of their reactions have shown that they contain specific functional groups. Chemically, the carbohydrates may be defined as optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. Some of the carbohydrates, which are sweet in taste, are also called sugars. The most common sugar, used in our homes is named as sucrose whereas the sugar present Reprint 2025-26 in milk is known as lactose. Carbohydrates are also called saccharides (Greek: sakcharon means sugar). Carbohydrates are classified on the basis of their behaviour on hydrolysis. They have been broadly divided into following three groups. 10.1.1 (i) Monosaccharides: A carbohydrate that cannot be hydrolysed further Classification of to give simpler unit of polyhydroxy aldehyde or ketone is called a Carbohydrates monosaccharide. About 20 monosaccharides are known to occur in nature. Some common examples are glucose, fructose, ribose, etc. (ii) Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides. They are further classified as disaccharides, trisaccharides, tetrasaccharides, etc., depending upon the number of monosaccharides, they provide on hydrolysis. Amongst these the most common are disaccharides. The two monosaccharide units obtained on hydrolysis of a disaccharide may be same or different. For example, one molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose whereas maltose gives two molecules of only glucose. (iii) Polysaccharides: Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Some common examples are starch, cellulose, glycogen, gums, etc. Polysaccharides are not sweet in taste, hence they are also called non-sugars. The carbohydrates may also be classified as either reducing or non- reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars. 10.1.2 Monosaccharides are further classified on the basis of number of carbon Monosaccharides atoms and the functional group present in them. If a monosaccharide contains an aldehyde group, it is known as an aldose and if it contains a keto group, it is known as a ketose. Number of carbon atoms constituting the monosaccharide is also introduced in the name as is evident from the examples given in Table 10.1 Table 10.1: Different Types of Monosaccharides Carbon atoms General term Aldehyde Ketone 3 Triose Aldotriose Ketotriose 4 Tetrose Aldotetrose Ketotetrose 5 Pentose Aldopentose Ketopentose 6 Hexose Aldohexose Ketohexose 7 Heptose Aldoheptose Ketoheptose 10.1.2.1 Glucose Glucose occurs freely in nature as well as in the combined form. It is present in sweet fruits and honey. Ripe grapes also contain glucose in large amounts. It is prepared as follows: Preparation of 1. From sucrose (Cane sugar): If sucrose is boiled with dilute HCl or Glucose H2SO4 in alcoholic solution, glucose and fructose are obtained in equal amounts. Chemistry 282 Reprint 2025-26 H+ C12 H 22 O11 + H 2 O → C 6 H12 O 6 + C 6 H12 O 6 Sucrose Glucose Fructose 2. From starch: Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute H2SO4 at 393 K under pressure. H + (C 6 H10 O 5 ) n + nH 2 O →393K; 2-3 atm nC 6 H12 O 6 Starch or cellulose Glucose Glucose is an aldohexose and is also known as dextrose. It is theStructure of monomer of many of the larger carbohydrates, namely starch, cellulose.Glucose It is probably the most abundant organic compound on earth. It was assigned the structure given below on the basis of the following evidences: CHO 1. Its molecular formula was found to be C6H12O6. 2. On prolonged heating with HI, it forms n-hexane, suggesting that all(CH OH )4 the six carbon atoms are linked in a straight chain. CH2OH Glucose 3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of a carbonyl group (>C = O) in glucose. 4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group. CHO COOH Br2 water (CHOH )4 (CH OH )4 CH2OH CH2OH Gluconic acid 5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms. 283 Biomolecules Reprint 2025-26 6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose. CHO COOH COOH Oxidation Oxidation (CHOH)4 (CHOH )4 (CHOH )4 CH 2OH COOH CH2 OH Saccharic Gluconic acid acid The exact spatial arrangement of different —OH groups was given by Fischer after studying many other properties. Its configuration is correctly represented as I. So gluconic acid is represented as II and saccharic acid as III. CHO COOH COOH H OH H OH H OH OH H OH H OH H H OH H OH H OH H OH H OH H OH CH2OH CH2OH COOH I II III Glucose is correctly named as D(+)-glucose. ‘D’ before the name of glucose represents the configuration whereas ‘(+)’ represents dextrorotatory nature of the molecule. It should be remembered that ‘D’ and ‘L’ have no relation with the optical activity of the compound. They are also not related to letter ‘d’ and ‘l’ (see Unit 6). The meaning of D– and L– notations is as follows. The letters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer of a compound with respect to configuration of some other compound, configuration of which is known. In the case of carbohydrates, this refers to their relation with a particular isomer of glyceraldehyde. Glyceraldehyde contains one asymmetric carbon atom and exists in two enantiomeric forms as shown below. (+) Isomer of glyceraldehyde has ‘D’ configuration. It means that when its structural formula is written on paper following specific conventions which you will study in higher classes, the –OH group lies on right hand side in the structure. All those compounds which can be chemically correlated to D (+) isomer of glyceraldehyde are said to have D- configuration whereas those which can be correlated to ‘L’ (–) isomer of glyceraldehyde are said to have L—configuration. In L (–) isomer –OH group is on left hand side as you can see in the structure. For assigning Chemistry 284 Reprint 2025-26 the configuration of monosaccharides, it is the lowest asymmetric carbon atom (as shown below) which is compared. As in (+) glucose, —OH on the lowest asymmetric carbon is on the right side which is comparable to (+) glyceraldehyde, so (+) glucose is assigned D-configuration. Other asymmetric carbon atoms of glucose are not considered for this comparison. Also, the structure of glucose and glyceraldehyde is written in a way that most oxidised carbon (in this case –CHO)is at the top. CHO H OH OH H CHO H OH H OH H OH CH2OH CH2OH D– (+) – Glyceraldehyde D–(+) – Glucose Cyclic The structure (I) of glucose explained most of its properties but the Structure following reactions and facts could not be explained by this structure. of Glucose 1. Despite having the aldehyde group, glucose does not give Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3. 2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group. 3. Glucose is found to exist in two different crystalline forms which are named as a and b. The a-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the b-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K. This behaviour could not be explained by the open chain structure (I) for glucose. It was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in ring formation. This explains the absence of —CHO group and also existence of glucose in two forms as shown below. These two cyclic forms exist in equilibrium with open chain structure. The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C1, called anomeric carbon 285 Biomolecules Reprint 2025-26 (the aldehyde carbon before cyclisation). Such isomers, i.e., a-form and b-form, are called anomers. The six membered cyclic structure of glucose is called pyranose structure (a– or b–), in analogy with pyran. Pyran is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring. The cyclic structure of glucose is more correctly represented by Haworth structure as given below. 10.1.2.2 Fructose Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide, sucrose. It is a natural monosaccharide found in fruits, honey and vegetables. In its pure form it is used as a sweetner. It is also an important ketohexose. Fructose also has the molecular formula C6H12O6 and Structure on the basis of its reactions it was found to contain a of Fructose ketonic functional group at carbon number 2 and six carbons in straight chain as in the case of glucose. It belongs to D-series and is a laevorotatory compound. It is appropriately written as D-(–)-fructose. Its open chain structure is as shown. It also exists in two cyclic forms which are obtained by the addition of —OH at C5 to the ( ) group. The ring, thus formed is a five membered ring and is named as furanose with analogy to the compound furan. Furan is a five membered cyclic compound with one oxygen and four carbon atoms. The cyclic structures of two anomers of fructose are represented by Haworth structures as given. Chemistry 286 Reprint 2025-26 10.1.3 You have already read that disaccharides on hydrolysis with dilute Disaccharides acids or enzymes yield two molecules of either the same or different monosaccharides. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage. In disaccharides, if the reducing groups of monosaccharides i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars, e.g., sucrose. On the other hand, sugars in which these functional groups are free, are called reducing sugars, for example, maltose and lactose. (i) Sucrose: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose. These two monosaccharides are held together by a glycosidic linkage between C1 of a-D-glucose and C2 of b-D-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non reducing sugar. Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–) and the product is named as invert sugar. (ii) Maltose: Another disaccharide, maltose is composed of two a-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II). The free aldehyde group can be produced at C1 of second glucose in solution and it shows reducing properties so it is a reducing sugar. 287 Biomolecules Reprint 2025-26 (iii) Lactose: It is more commonly known as milk sugar since this disaccharide is found in milk. It is composed of b-D-galactose and b-D-glucose. The linkage is between C1 of galactose and C4 of glucose. Free aldehyde group may be produced at C-1 of glucose unit, hence it is also a reducing sugar. 10.1.4 Polysaccharides Polysaccharides contain a large number of monosaccharide units joined together by glycosidic linkages. These are the most commonly encountered carbohydrates in nature. They mainly act as the food storage or structural materials. (i) Starch: Starch is the main storage polysaccharide of plants. It is the most important dietary source for human beings. High content of starch is found in cereals, roots, tubers and some vegetables. It is a polymer of a-glucose and consists of two components— Amylose and Amylopectin. Amylose is water soluble component which constitutes about 15-20% of starch. Chemically amylose is a long unbranched chain with 200-1000 a-D-(+)-glucose units held together by C1– C4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80- 85% of starch. It is a branched chain polymer of a-D-glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage. Chemistry 288 Reprint 2025-26 (ii) Cellulose: Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain polysaccharide composed only of b-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit. (iii) Glycogen: The carbohydrates are stored in animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin and is rather more highly branched. It is present in liver, muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose. Glycogen is also found in yeast and fungi. 10.1.5 Carbohydrates are essential for life in both plants and animals. They Importance of form a major portion of our food. Honey has been used for a long time Carbohydrates as an instant source of energy by ‘Vaids’ in ayurvedic system of medicine. Carbohydrates are used as storage molecules as starch in plants and glycogen in animals. Cell wall of bacteria and plants is made up of cellulose. We build furniture, etc. from cellulose in the form 289 Biomolecules Reprint 2025-26 of wood and clothe ourselves with cellulose in the form of cotton fibre. They provide raw materials for many important industries like textiles, paper, lacquers and breweries. Two aldopentoses viz. D-ribose and 2-deoxy-D-ribose (Section 10.5.1, Class XII) are present in nucleic acids. Carbohydrates are found in biosystem in combination with many proteins and lipids. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 10.1 Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. 10.2 What are the expected products of hydrolysis of lactose? 10.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? 10.210.210.210.210.2 ProteinsProteinsProteinsProteinsProteins Proteins are the most abundant biomolecules of the living system. Chief sources of proteins are milk, cheese, pulses, peanuts, fish, meat, etc. They occur in every part of the body and form the fundamental basis of structure and functions of life. They are also required for growth and maintenance of body. The word protein is derived from Greek word, “proteios” which means primary or of prime importance. All proteins are polymers of a-amino acids. 10.2.1 Amino Amino acids contain amino (–NH2) and carboxyl (–COOH) functional groups. Depending upon the relative position of amino group with Acids respect to carboxyl group, the amino acids can be R CH COOH classified as a, b, g, d and so on. Only a-amino acids are obtained on hydrolysis of proteins. They NH2 may contain other functional groups also. a-amino acid All a-amino acids have trivial names, which (R = side chain) usually reflect the property of that compound or its source. Glycine is so named since it has sweet taste (in Greek glykos means sweet) and tyrosine was first obtained from cheese (in Greek, tyros means cheese.) Amino acids are generally represented by a three letter symbol, sometimes one letter symbol is also used. Structures of some commonly occurring amino acids along with their 3-letter and 1-letter symbols are given in Table 10.2. COOH Table 10.2: Natural Amino Acids H2N H R Name of the Characteristic feature Three letter One letter amino acids of side chain, R symbol code 1. Glycine H Gly G 2. Alanine – CH3 Ala A 3. Valine* (H3C)2CH- Val V 4. Leucine* (H3C)2CH-CH2- Leu L Chemistry 290 Reprint 2025-26 5. Isoleucine* H3C-CH2-CH- Ile I | CH3 6. Arginine* HN=C-NH-(CH2)3- Arg R | NH2 7. Lysine* H2N-(CH2)4- Lys K 8. Glutamic acid HOOC-CH2-CH2- Glu E 9. Aspartic acid HOOC-CH2- Asp D O || 10. Glutamine H2N-C-CH2-CH2- Gln Q O || 11. Asparagine H2N-C-CH2- Asn N 12. Threonine* H3C-CHOH- Thr T 13. Serine HO-CH2- Ser S 14. Cysteine HS-CH2- Cys C 15. Methionine* H3C-S-CH2-CH2- Met M 16. Phenylalanine* C6H5-CH2- Phe F 17. Tyrosine (p)HO-C6H4-CH2- Tyr Y –CH2 18. Tryptophan* Trp W N H 19. Histidine* His H 20. Proline Pro P * essential amino acid, a = entire structure 10.2.2 Amino acids are classified as acidic, basic or neutral depending upon Classification of the relative number of amino and carboxyl groups in their molecule. Amino Acids Equal number of amino and carboxyl groups makes it neutral; more number of amino than carboxyl groups makes it basic and more carboxyl groups as compared to amino groups makes it acidic. The amino acids, which can be synthesised in the body, are known as non- essential amino acids. On the other hand, those which cannot be synthesised in the body and must be obtained through diet, are known as essential amino acids (marked with asterisk in Table 10.2). 291 Biomolecules Reprint 2025-26 Amino acids are usually colourless, crystalline solids. These are water-soluble, high melting solids and behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges. In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids and bases. Except glycine, all other naturally occurring a-amino acids are optically active, since the a-carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms. Most naturally occurring amino acids have L-configuration. L-Aminoacids are represented by writing the –NH2 group on left hand side. 10.2.3 Structure You have already read that proteins are the polymers of a-amino acids of Proteins and they are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between –COOH group and –NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other. This results in the elimination of a water molecule and formation of a peptide bond –CO–NH–. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine. If a third amino acid combines to a dipeptide, the product is called a tripeptide. A tripeptide contains three amino acids linked by two peptide linkages. Similarly when four, five or six amino acids are linked, the respective products are known as tetrapeptide, pentapeptide or hexapeptide, respectively. When the number of such amino acids is more than ten, then the products are called polypeptides. A polypeptide with more than hundred amino acid residues, having molecular mass higher than 10,000u is called a protein. However, the distinction between a polypeptide and a protein is not very sharp. Polypeptides with fewer amino acids are likely to be called proteins if they ordinarily have a well defined conformation of a protein such as insulin which contains 51 amino acids. Proteins can be classified into two types on the basis of their molecular shape. (a) Fibrous proteins When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, then fibre– like structure is formed. Such proteins are generally insoluble in water. Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc. Chemistry 292 Reprint 2025-26 (b) Globular proteins This structure results when the chains of polypeptides coil around to give a spherical shape. These are usually soluble in water. Insulin and albumins are the common examples of globular proteins. Structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quaternary, each level being more complex than the previous one. (i) Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e., the sequence of amino acids creates a different protein. (ii) Secondary structure of proteins: The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures viz. a-helix and b-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between and –NH– groups of the peptide bond. a-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with the Fig. 10.1: a-Helix –NH group of each amino acid residue hydrogen bonded to the structure of proteins C O of an adjacent turn of the helix as shown in Fig.10.1. In b-pleated sheet structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as b-pleated sheet. (iii) Tertiary structure of proteins: The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous and globular. The main forces which stabilise the 2° and 3° structures of proteins are hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction. Fig. 10.2: b-Pleated sheet structure of (iv) Quaternary structure of proteins: Some of the proteins proteins are composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other is known as quaternary structure. 293 Biomolecules Reprint 2025-26 A diagrammatic representation of all these four structures is given in Figure 10.3 where each coloured ball represents an amino acid. Fig. 10.3: Diagrammatic representation of protein structure (two sub-units of two types in quaternary structure) Fig. 10.4: Primary, secondary, tertiary and quaternary structures of haemoglobin 10.2.4 Protein found in a biological system with a unique three-dimensional Denaturation of structure and biological activity is called a native protein. When a Proteins protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of Chemistry 294 Reprint 2025-26 protein. During denaturation secondary and tertiary structures are destroyed but primary structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 10.4 The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain. 10.5 Where does the water present in the egg go after boiling the egg? 10.310.310.310.310.3 EnzymesEnzymesEnzymesEnzymesEnzymes Life is possible due to the coordination of various chemical reactions in living organisms. An example is the digestion of food, absorption of appropriate molecules and ultimately production of energy. This process involves a sequence of reactions and all these reactions occur in the body under very mild conditions. This occurs with the help of certain biocatalysts called enzymes. Almost all the enzymes are globular proteins. Enzymes are very specific for a particular reaction and for a particular substrate. They are generally named after the compound or class of compounds upon which they work. For example, the enzyme that catalyses hydrolysis of maltose into glucose is named as maltase. Maltase C12 H 22 O11  2 C 6 H12 O 6 Maltose G lucose Sometimes enzymes are also named after the reaction, where they are used. For example, the enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate are named as oxidoreductase enzymes. The ending of the name of an enzyme is -ase. 10.3.1 Mechanism Enzymes are needed only in small quantities for the progress of a reaction. of Enzyme Similar to the action of chemical catalysts, enzymes are said to reduce Action the magnitude of activation energy. For example, activation energy for acid hydrolysis of sucrose is 6.22 kJ mol–1, while the activation energy is only 2.15 kJ mol–1 when hydrolysed by the enzyme, sucrase. Mechanism for the enzyme action has been discussed. 10.410.410.410.410.4 VitaminsVitaminsVitaminsVitaminsVitamins It has been observed that certain organic compounds are required in small amounts in our diet but their deficiency causes specific diseases. These compounds are called vitamins. Most of the vitamins cannot be synthesised in our body but plants can synthesise almost all of them, so they are considered as essential food factors. However, the bacteria of the gut can produce some of the vitamins required by us. All the vitamins are generally available in our diet. Different vitamins belong to various chemical classes and it is difficult to define them on the basis of structure. They are generally regarded as organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth 295 Biomolecules Reprint 2025-26 and health of the organism. Vitamins are designated by alphabets A, B, C, D, etc. Some of them are further named as sub-groups e.g. B1, B2, B6, B12, etc. Excess of vitamins is also harmful and vitamin pills should not be taken without the advice of doctor. The term “Vitamine” was coined from the word vital + amine since the earlier identified compounds had amino groups. Later work showed that most of them did not contain amino groups, so the letter ‘e’ was dropped and the term vitamin is used these days. 10.4.1 Vitamins are classified into two groups depending upon their solubility Classification of in water or fat. Vitamins (i) Fat soluble vitamins: Vitamins which are soluble in fat and oils but insoluble in water are kept in this group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. (ii) Water soluble vitamins: B group vitamins and vitamin C are soluble in water so they are grouped together. Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body. Some important vitamins, their sources and diseases caused by their deficiency are listed in Table 10.3. Table 10.3: Some important Vitamins, their Sources and their Deficiency Diseases Sl. Name of Sources Deficiency diseases No. Vitamins 1. Vitamin A Fish liver oil, carrots, X e r o p h t h a l m i a butter and milk (hardening of cornea of eye) Night blindness 2. Vitamin B1 Yeast, milk, green Beri beri (loss of appe- (Thiamine) vegetables and cereals tite, retarded growth) 3. Vitamin B2 Milk, eggwhite, liver, Cheilosis (fissuring at (Riboflavin) kidney corners of mouth and lips), digestive disorders and burning sensation of the skin. 4. Vitamin B6 Yeast, milk, egg yolk, Convulsions (Pyridoxine) cereals and grams 5. Vitamin B12 Meat, fish, egg and Pernicious anaemia curd (RBC deficient in haemoglobin) 6. Vitamin C Citrus fruits, amla and Scurvy (bleeding gums) (Ascorbic acid) green leafy vegetables 7. Vitamin D Exposure to sunlight, Rickets (bone deformities fish and egg yolk in children) and osteo- malacia (soft bones and joint pain in adults) Chemistry 296 Reprint 2025-26 8. Vitamin E Vegetable oils like wheat Increased fragility of germ oil, sunflower oil, RBCs and muscular etc. weakness 9. Vitamin K Green leafy vegetables Increased blood clotting time 10.5.5.5.5.5 NucleicNucleicNucleicNucleicNucleic AcidsAcidsAcidsAcidsAcids Every generation of each and every species resembles its ancestors in many ways. How are these characteristics transmitted from one generation to the next? It has been observed that nucleus of a living cell is responsible for this transmission of inherent characters, also called heredity. The particles in nucleus of the cell, responsible for heredity, are called chromosomes which are made up of proteins and another type of biomolecules called nucleic acids. These are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Since nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides. James Dewey Watson Born in Chicago, Illinois, in 1928, Dr Watson received his Ph.D. (1950) from Indiana University in Zoology. He is best known for his discovery of the structure of DNA for which he shared with Francis Crick and Maurice Wilkins the 1962 Nobel prize in Physiology and Medicine. They proposed that DNA molecule takes the shape of a double helix, an elegantly simple structure that resembles a gently twisted ladder. The rails of the ladder are made of alternating units of phosphate and the sugar deoxyribose; the rungs are each composed of a pair of purine/ pyrimidine bases. This research laid the foundation for the emerging field of molecular biology. The complementary pairing of nucleotide bases explains how identical copies of parental DNA pass on to two daughter cells. This research launched a revolution in biology that led to modern recombinant DNA techniques. 10.5.1 Chemical Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric Composition acid and nitrogen containing heterocyclic compounds (called bases). In of Nucleic DNA molecules, the sugar moiety is b-D-2-deoxyribose whereas in Acids RNA molecule, it is b-D-ribose. 297 Biomolecules Reprint 2025-26 DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also contains four bases, the first three bases are same as in DNA but the fourth one is uracil (U). Cytosine (C) Thymine (T) Uracil (U) 10.5.2 Structure A unit formed by the attachment of a base to 1¢ position of sugar is of Nucleic known as nucleoside. In nucleosides, the sugar carbons are numbered Acids as 1¢, 2¢, 3¢, etc. in order to distinguish these from the bases (Fig. 10.5a). When nucleoside is linked to phosphoric acid at 5¢-position of sugar moiety, we get a nucleotide (Fig. 10.5). Fig. 10.5: Structure of (a) a nucleoside and (b) a nucleotide Nucleotides are joined together by phosphodiester linkage between 5¢ and 3¢ carbon atoms of the pentose sugar. The formation of a typical dinucleotide is shown in Fig. 10.6. Chemistry 298 Reprint 2025-26 Fig. 10.6: Formation of a dinucleotide A simplified version of nucleic acid chain is as shown below. Base Base Base Sugar Phosphate Sugar Phosphate Sugar n Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure also. James Watson and Francis Crick gave a double strand helix structure for DNA (Fig. 10.7). Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine. In secondary structure of RNA single stranded helics is present which sometimes foldsback on itself. RNA molecules are of three types and they perform different functions. They are named as messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA). Fig. 10.7: Double strand helix structure for DNA 299 Biomolecules Reprint 2025-26 Har Gobind Khorana Har Gobind Khorana, was born in 1922. He obtained his M.Sc. degree from Punjab University in Lahore. He worked with Professor Vladimir Prelog, who moulded Khorana’s thought and philosophy towards science, work and effort. After a brief stay in India in 1949, Khorana went back to England and worked with Professor G.W. Kenner and Professor A.R.Todd. It was at Cambridge, U.K. that he got interested in both proteins and nucleic acids. Dr Khorana shared the Nobel Prize for Medicine and Physiology in 1968 with Marshall Nirenberg and Robert Holley for cracking the genetic code. DNA Fingerprinting It is known that every individual has unique fingerprints. These occur at the tips of the fingers and have been used for identification for a long time but these can be altered by surgery. A sequence of bases on DNA is also unique for a person and information regarding this is called DNA fingerprinting. It is same for every cell and cannot be altered by any known treatment. DNA fingerprinting is now used (i) in forensic laboratories for identification of criminals. (ii) to determine paternity of an individual. (iii) to identify the dead bodies in any accident by comparing the DNA’s of parents or children. (iv) to identify racial groups to rewrite biological evolution. 10.5.3 Biological DNA is the chemical basis of heredity and may be regarded as the reserve Functions of genetic information. DNA is exclusively responsible for maintaining of Nucleic the identity of different species of organisms over millions of years. A Acids DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells. Another important function of nucleic acids is the protein synthesis in the cell. Actually, the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA. 10.610.610.610.610.6 HormonesHormonesHormonesHormonesHormones Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly in the blood stream which transports them to the site of action. In terms of chemical nature, some of these are steroids, e.g., estrogens and androgens; some are poly peptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine. Hormones have several functions in the body. They help to maintain the balance of biological activities in the body. The role of insulin in keeping the blood glucose level within the narrow limit is an example of this function. Insulin is released in response to the rapid rise in blood glucose level. On the other hand hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood. Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex hormones play role in growth and development. Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine. Abnormally low level of thyroxine leads Chemistry 300 Reprint 2025-26 to hypothyroidism which is characterised by lethargyness and obesity. Increased level of thyroxine causes hyperthyroidism. Low level of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This condition is largely being controlled by adding sodium iodide to commercial table salt (“Iodised” salt). Steroid hormones are produced by adrenal cortex and gonads (testes in males and ovaries in females). Hormones released by the adrenal cortex play very important role in the functions of the body. For example, glucocorticoids control the carbohydrate metabolism, modulate inflammatory reactions, and are involved in reactions to stress. The mineralocorticoids control the level of excretion of water and salt by the kidney. If adrenal cortex does not function properly then one of the results may be Addison’s disease characterised by hypoglycemia, weakness and increased susceptibility to stress. The disease is fatal unless it is treated by glucocorticoids and mineralocorticoids. Hormones released by gonads are responsible for development of secondary sex characters. Testosterone is the major sex hormone produced in males. It is responsible for development of secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is the main female sex hormone. It is responsible for development of secondary female characteristics and participates in the control of menstrual cycle. Progesterone is responsible for preparing the uterus for implantation of fertilised egg. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 10.6 Why cannot vitamin C be stored in our body? 10.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed? 10.8 When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA? SummarySummarySummarySummarySummary Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules which provide such units on hydrolysis. They are broadly classified into three groups — monosaccharides, disaccharides and polysaccharides. Glucose, the most important source of energy for mammals, is obtained by the digestion of starch. Monosaccharides are held together by glycosidic linkages to form disaccharides or polysaccharides. Proteins are the polymers of about twenty different a-amino acids which are linked by peptide bonds. Ten amino acids are called essential amino acids because they cannot be synthesised by our body, hence must be provided through diet. Proteins perform various structural and dynamic functions in the organisms. Proteins which contain only a-amino acids are called simple proteins. The secondary or tertiary structure of proteins get disturbed on change of pH or temperature and they are not able to perform their functions. This is called denaturation of proteins. Enzymes are biocatalysts which speed up the reactions in biosystems. They are very specific and selective in their action and chemically majority of enzymes are proteins. Vitamins are accessory food factors required in the diet. They are classified as fat soluble (A, D, E and K) and water soluble (B group and C). Deficiency of vitamins leads to many diseases. 301 Biomolecules Reprint 2025-26 Nucleic acids are the polymers of nucleotides which in turn consist of a base, a pentose sugar and phosphate moiety. Nucleic acids are responsible for the transfer of characters from parents to offsprings. There are two types of nucleic acids — DNA and RNA. DNA contains a five carbon sugar molecule called 2-deoxyribose whereas RNA contains ribose. Both DNA and RNA contain adenine, guanine and cytosine. The fourth base is thymine in DNA and uracil in RNA. The structure of DNA is a double strand whereas RNA is a single strand molecule. DNA is the chemical basis of heredity and have the coded message for proteins to be synthesised in the cell. There are three types of RNA — mRNA, rRNA and tRNA which actually carry out the protein synthesis in the cell. ExercisesExercisesExercisesExercisesExercises 10.1 What are monosaccharides? 10.2 What are reducing sugars? 10.3 Write two main functions of carbohydrates in plants. 10.4 Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose. 10.5 What do you understand by the term glycosidic linkage? 10.6 What is glycogen? How is it different from starch? 10.7 What are the hydrolysis products of (i) sucrose and (ii) lactose? 10.8 What is the basic structural difference between starch and cellulose? 10.9 What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO3 10.10 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure. 10.11 What are essential and non-essential amino acids? Give two examples of each type. 10.12 Define the following as related to proteins (i) Peptide linkage (ii) Primary structure (iii) Denaturation. 10.13 What are the common types of secondary structure of proteins? 10.14 What type of bonding helps in stabilising the a-helix structure of proteins? 10.15 Differentiate between globular and fibrous proteins. 10.16 How do you explain the amphoteric behaviour of amino acids? 10.17 What are enzymes? 10.18 What is the effect of denaturation on the structure of proteins? 10.19 How are vitamins classified? Name the vitamin responsible for the coagulation of blood. 10.20 Why are vitamin A and vitamin C essential to us? Give their important sources. 10.21 What are nucleic acids? Mention their two important functions. 10.22 What is the difference between a nucleoside and a nucleotide? 10.23 The two strands in DNA are not identical but are complementary. Explain. 10.24 Write the important structural and functional differences between DNA and RNA. 10.25 What are the different types of RNA found in the cell? Chemistry 302 Reprint 2025-26

9.3 Alkenes Alkenes are unsaturated hydrocarbons Fig. 9.4 Orbital picture of ethene depictingcontaining at least one double bond. What σ bonds only should be the general formula of alkenes? If there is one double bond between two carbon 9.3.2 Nomenclature atoms in alkenes, they must possess two For nomenclature of alkenes in IUPAC system, hydrogen atoms less than alkanes. Hence, the longest chain of carbon atoms containing general formula for alkenes is CnH2n. Alkenes the double bond is selected. Numbering of the are also known as olefins (oil forming) since chain is done from the end which is nearer to Reprint 2025-26 Hydrocarbons 307 Fig. 9.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles and bond lengths the double bond. The suffix ‘ene’ replaces ‘ane’ Solutionof alkanes. It may be remembered that first member of alkene series is: CH2 (replacing (i) 2,8-Dimethyl-3, 6-decadiene; n by 1 in CnH2n) known as methene but has (ii) 1,3,5,7 Octatetraene; a very short life. As already mentioned, first (iii) 2-n-Propylpent-1-ene; stable member of alkene series is C2H4 known (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene;as ethylene (common) or ethene (IUPAC). IUPAC names of a few members of alkenes Problem 9.8 are given below : Calculate number of sigma (σ) and pi (π) Structure IUPAC name bonds in the above structures (i-iv). CH3 – CH = CH2 Propene SolutionCH3 – CH2 – CH = CH2 But – l - ene σ bonds : 33, π bonds : 2CH3 – CH = CH–CH3 But-2-ene σ bonds : 17, π bonds : 4CH2 = CH – CH = CH2 Buta – 1,3 - diene σ bonds : 23, π bond : 1CH2 = C – CH3 2-Methylprop-1-ene | σ bonds : 41, π bond : 1 CH3 CH2 = CH – CH – CH3 3-Methylbut-1-ene 9.3.3 Isomerism | Alkenes show both structural isomerism and CH3 geometrical isomerism. Structural isomerism : As in alkanes, ethene Problem 9.7 (C2H4) and propene (C3H6) can have only one Write IUPAC names of the following structure but alkenes higher than propene compounds: have different structures. Alkenes possessing (i) (CH3)2CH – CH = CH – CH2 – CH C4H8 as molecular formula can be written in  the following three ways: CH3 – CH – CH | I. 1 2 3 4 C2H5 CH2 = CH – CH2 – CH3 (ii) But-1-ene (C4H8) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 II. 1 2 3 4 | | CH3 – CH = CH – CH3 CH3 – CHCH = C – CH2 – CHCH3 | But-2-ene CH3 (C4H8) Reprint 2025-26 308 chemistry III. 1 2 3 In (a), the two identical atoms i.e., both CH2 = C – CH3 the X or both the Y lie on the same side | of the double bond but in (b) the two X or CH3 two Y lie across the double bond or on the 2-Methylprop-1-ene opposite sides of the double bond. This (C4H8) results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in Structures I and III, and II and III are the two arrangements is different. Therefore,the examples of chain isomerism whereas they are stereoisomers. They would have thestructures I and II are position isomers. same geometry if atoms or groups around C=C bond can be rotated but rotation around Problem 9.9 C=C bond is not free. It is restricted. For Write structures and IUPAC names of understanding this concept, take two pieces different structural isomers of alkenes of strong cardboards and join them with the corresponding to C5H10. help of two nails. Hold one cardboard in your Solution one hand and try to rotate the other. Can (a) CH2 = CH – CH2 – CH2 – CH3 you really rotate the other cardboard ? The answer is no. The rotation is restricted. This Pent-1-ene illustrates that the restricted rotation of atoms (b) CH3 – CH=CH – CH2 – CH3 or groups around the doubly bonded carbon Pent-2-ene atoms gives rise to different geometries of such compounds. The stereoisomers of this (c) CH3 – C = CH – CH3 type are called geometrical isomers. The | isomer of the type (a), in which two identical CH3 atoms or groups lie on the same side of the 2-Methylbut-2-ene double bond is called cis isomer and the (d) CH3 – CH – CH = CH2 other isomer of the type (b), in which identical | atoms or groups lie on the opposite sides of CH3 the double bond is called trans isomer . Thus 3-Methylbut-1-ene cis and trans isomers have the same structure but have different configuration (arrangement (e) CH2 = C – CH2 – CH3 of atoms or groups in space). Due to different | arrangement of atoms or groups in space, CH3 these isomers differ in their properties like 2-Methylbut-1-ene melting point, boiling point, dipole moment, solubility etc. Geometrical or cis-trans isomersGeometrical isomerism: Doubly bonded of but-2-ene are represented below :carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that Reprint 2025-26 Hydrocarbons 309 trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2 understood by drawing geometries of the two (iii) C6H5CH = CH – CH3forms as given below from which it is clear that in the trans-but-2-ene, the two methyl (iv) CH3CH = CCl CH3 groups are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus Solution making the trans form non-polar. (iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom. 9.3.4 Preparation 1. From alkynes: Alkynes on partial reduction with calculated amount of cis-But-2-ene trans-But-2-ene dihydrogen in the presence of palladised (µ = 0.33D) (µ = 0) charcoal partially deactivated with poisons like sulphur compounds or quinoline give In the case of solids, it is observed that the alkenes. Partially deactivated palladisedtrans isomer has higher melting point than charcoal is known as Lindlar’s catalyst.the cis form. Alkenes thus obtained are having cis Geometrical or cis-trans isomerism geometry. However, alkynes on reductionis also shown by alkenes of the types with sodium in liquid ammonia form transXYC = CXZ and XYC = CZW alkenes. Problem 9.10 Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (9.30) (ii) C2H5CCH3 = CCH3C2H5 Solution (9.31) iii) CH≡ CH+H2 Pd/C CH2 =CH2 (9.32) Ethyne Ethene CH3–C≡ CH+H2 Pd/C CH3–CH =CH2 iv) Propyne Propene (9.33) Will propene thus obtained show Problem 9.11 geometrical isomerism? Think for the reason in support of your answer. Which of the following compounds will show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) on (i) (CH3)2C = CH – C2H5 heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, Reprint 2025-26 310 chemistry ethanol) eliminate one molecule of halogen takes out one hydrogen atom from the acid to form alkenes. This reaction is β-carbon atom. known as dehydrohalogenation i.e., removal of halogen acid. This is example of β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). (9.37) 9.3.5 Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. (9.34) The first three members are gases, the next fourteen are liquids and the higher ones are Nature of halogen atom and the alkyl group solids. Ethene is a colourless gas with a faint determine rate of the reaction. It is observed sweet smell. All other alkenes are colourless that for halogens, the rate is: iodine > and odourless, insoluble in water but fairly bromine > chlorine, while for alkyl groups soluble in non-polar solvents like benzene, it is : tertiary > secondary > primary. petroleum ether. They show a regular increase 3. From vicinal dihalides: Dihalides in in boiling point with increase in size i.e., every which two halogen atoms are attached – CH2 group added increases boiling point by to two adjacent carbon atoms are known 20–30 K. Like alkanes, straight chain alkenes as vicinal dihalides. Vicinal dihalides on have higher boiling point than isomeric treatment with zinc metal lose a molecule branched chain compounds. of ZnX2 to form an alkene. This reaction Chemical properties is known as dehalogenation. Alkenes are the rich source of loosely held CH2Br–CH2Br + Zn CH2=CH2+ ZnBr2 pi (π) electrons, due to which they show (9.35) addition reactions in which the electrophiles add on to the carbon-carbon double bond toCH3CHBr–CH2Br + Zn CH3CH=CH2 form the addition products. Some reagents +ZnBr2 also add by free radical mechanism. There (9.36) are cases when under special conditions, alkenes also undergo free radical substitution4. From alcohols by acidic dehydration: reactions. Oxidation and ozonolysis reactions You have read during nomenclature of are also quite prominent in alkenes. A brief different homologous series in Unit 12 description of different reactions of alkenes that alcohols are the hydroxy derivatives is given below: of alkanes. They are represented by R–OH where, R is CnH2n+1. Alcohols on heating 1. Addition of dihydrogen: Alkenes add with concentrated sulphuric acid form up one molecule of dihydrogen gas in alkenes with the elimination of one water the presence of finely divided nickel, molecule. Since a water molecule is palladium or platinum to form alkanes eliminated from the alcohol molecule in (Section 9.2.2) the presence of an acid, this reaction is 2. Addition of halogens : Halogens like known as acidic dehydration of alcohols. bromine or chlorine add up to alkene to This reaction is also the example of form vicinal dihalides. However, iodine β-elimination reaction since –OH group does not show addition reaction under Reprint 2025-26 Hydrocarbons 311 normal conditions. The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction involving cyclic halonium ion formation (9.42) which you will study in higher classes. Markovnikov, a Russian chemist made a generalisation in 1869 after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that negative part of the addendum (adding (9.38) molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Thus according to this rule, product I i.e., 2-bromopropane is expected. In actual practice, this is the principal product of the reaction. This generalisation of Markovnikov (9.39) rule can be better understood in terms of mechanism of the reaction. 3. Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes Mechanism to form alkyl halides. The order of Hydrogen bromide provides an electrophile, reactivity of the hydrogen halides is H +, which attacks the double bond to form HI > HBr > HCl. Like addition of halogens carbocation as shown below : to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical alkenes (a) less stable (b) more stableAddition reactions of HBr to symmetrical primary carbocation secondary carbocationalkenes (similar groups attached to double bond) take place by electrophilic addition (i) The secondary carbocation (b) is more mechanism. stable than the primary carbocation (a), therefore, the former predominates CH2=CH2+H–Br CH3–CH2–Br (9.40) because it is formed at a faster rate. (ii) The carbocation (b) is attacked by Br– ionCH3–CH=CH–CH3+HBr CH3–CH–CHCH3 to form the product as follows : Br (9.41) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two 2-Bromopropane possible products are I and II. (major product) Reprint 2025-26 312 chemistry Anti Markovnikov addition or peroxide effect or Kharash effect In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to the Markovnikov rule. This happens only with HBr but not with HCl The secondary free radical obtained in the and Hl. This addition reaction was observed above mechanism (step iii) is more stable than by M.S. Kharash and F.R. Mayo in 1933 the primary. This explains the formation of at the University of Chicago. This reaction 1-bromopropane as the major product. It may is known as peroxide or Kharash effect be noted that the peroxide effect is not observed or addition reaction anti to Markovnikov in addition of HCl and HI. This may be due rule. to the fact that the H–Cl bond being (C6H5CO)2O2 stronger (430.5 kJ mol –1) than H–Br bond CH3 – CH=CH2+HBr CH3–CH2 (363.7 kJ mol –1), is not cleaved by the free radical, whereas the H–I bond is weaker CH2Br (296.8 kJ mol –1) and iodine free radicals 1–Bromopropane combine to form iodine molecules instead of adding to the double bond. (9.43) Mechanism : Peroxide effect proceeds via Problem 9.12 free radical chain mechanism as given below: Write IUPAC names of the products obtained by addition reactions of HBr to(i) hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Solution Homolysis C. 6H5+H–Br C6H3+ B. r(ii) 4. Addition of sulphuric acid : Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction. Reprint 2025-26 Hydrocarbons 313 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (9.49) KMnO4/H+ CH3 – CH=CH–CH3 2CH3COOH (9.44) But-2-ene Ethanoic acid (9.50) 7. Ozonolysis : Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H2O to smaller molecules. This reaction is highly useful in detecting the position of the double (9.45) bond in alkenes or other unsaturated compounds.5. Addition of water : In the presence of a few drops of concentrated sulphuric acid alkenes react with water to form alcohols, in accordance with the Markovnikov rule. (9.51) (9.46) 6. Oxidation: Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO4 solution is used as a test for unsaturation. (9.52) 8. Polymerisation: You are familiar with (9.47) polythene bags and polythene sheets. Polythene is obtained by the combination of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules thus obtained are called (9.48) polymers. This reaction is known as b) Acidic potassium permanganate or acidic polymerisation. The simple compounds potassium dichromate oxidises alkenes to from which polymers are made are called Reprint 2025-26 314 chemistry monomers. Other alkenes also undergo are named as derivatives of the corresponding polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. n(CH2 =CH2) High temp./pressureCatalyst —( CH2–CH2 )— The position of the triple bond is indicated by the first triply bonded carbon. Common Polythene and IUPAC names of a few members of alkyne (9.53) series are given in Table 9.2. High temp./pressure You have already learnt that ethyne and n(CH3 –CH=CH2) Catalyst —( CH–CH2 )—n propyne have got only one structure but there are two possible structures for butyne – CH3 (i) but-1-yne and (ii) but-2-yne. Since these Polypropene two compounds differ in their structures (9.54) due to the position of the triple bond, they Polymers are used for the manufacture of plastic are known as position isomers. In how bags, squeeze bottles, refrigerator dishes, toys, many ways, you can construct the structure pipes, radio and T.V. cabinets etc. Polypropene for the next homologue i.e., the next alkyne is used for the manufacture of milk crates, with molecular formula C5H8? Let us try to plastic buckets and other moulded articles. arrange five carbon atoms with a continuous Though these materials have now become chain and with a side chain. Following are the common, excessive use of polythene and possible structures : polypropylene is a matter of great concern for Structure IUPAC name all of us. 1 2 3 4 5 I. HC≡ C– CH2– CH2– CH3 Pent–1-yne

9.7 (c)]. This can be represented in the form This is often referred to as Hückel Rule.of two doughtnuts (rings) of electron clouds [Fig. 9.7 (d)], one above and one below the Some examples of aromatic compounds are plane of the hexagonal ring as shown below: given below: (electron cloud) Fig. 9.7 (c) Fig. 9.7 (d) The six π electrons are thus delocalised and can move freely about the six carbon nuclei, instead of any two as shown in Fig. 9.6 (a) or (b). The delocalised π electron cloud is attracted more strongly by the nuclei of the carbon atoms than the electron cloud localised between two carbon atoms. Therefore, presence of delocalised π electrons in benzene makes it more stable than the hypothetical cyclohexatriene. X-Ray diffraction data reveals that benzene is a planar molecule. Had any one of the above structures of benzene (A or B) been correct, 9.5.4 Preparation of Benzene two types of C—C bond lengths were expected. Benzene is commercially isolated from coalHowever, X-ray data indicates that all the tar. However, it may be prepared in thesix C—C bond lengths are of the same order laboratory by the following methods.(139 pm) which is intermediate between C— C single bond (154 pm) and C—C double (i) Cyclic polymerisation of ethyne: bond (133 pm). Thus the absence of pure (Section 9.4.4) double bond in benzene accounts for the (ii) Decarboxylation of aromatic acids: reluctance of benzene to show addition Sodium salt of benzoic acid on heating reactions under normal conditions, thus with sodalime gives benzene. explaining the unusual behaviour of benzene. 9.5.3 Aromaticity Benzene was considered as parent ‘aromatic’ compound. Now, the name is applied to all the ring systems whether or not having benzene (9.70) ring, possessing following characteristics. Reprint 2025-26 322 chemistry (iii) Reduction of phenol: Phenol is reduced (ii) Halogenation: Arenes react with halogens to benzene by passing its vapours over in the presence of a Lewis acid like anhydrous heated zinc dust FeCl3, FeBr3 or AlCl3 to yield haloarenes. (9.71) Chlorobenzene 9.5.5 Properties (9.73) (iii) Sulphonation: The replacement of aPhysical properties hydrogen atom by a sulphonic acid group inAromatic hydrocarbons are non- polar a ring is called sulphonation. It is carried outmolecules and are usually colourless liquids by heating benzene with fuming sulphuricor solids with a characteristic aroma. You are acid (oleum).also familiar with naphthalene balls which are used in toilets and for preservation of clothes because of unique smell of the compound and the moth repellent property. Aromatic hydrocarbons are immiscible with water but are readily miscible with organic solvents. They burn with sooty flame. Chemical properties (9.74) Arenes are characterised by electrophilic (iv) Friedel-Crafts alkylation reaction:substitution reactions. However, under When benzene is treated with an alkyl halidespecial conditions they can also undergo in the presence of anhydrous aluminiumaddition and oxidation reactions. chloride, alkylbenene is formed. Electrophilic substitution reactions The common electrophilic substitution reactions of arenes are nitration, halogenation, sulphonation, Friedel Craft’s alkylation and acylation reactions in which attacking reagent is an electrophile (E +) (i) Nitration: A nitro group is introduced (9.75) into benzene ring when benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture). (9.76) Why do we get isopropyl benzene on treating benzene with 1-chloropropane instead of n-propyl benzene? (v) Friedel-Crafts acylation reaction: The reaction of benzene with an acyl halide or (9.72) acid anhydride in the presence of Lewis acids (AlCl3) yields acyl benzene. Nitrobenzene Reprint 2025-26 Hydrocarbons 323 (9.77) In the case of nitration, the electrophile, nitronium ion, is produced by transfer of a proton (from sulphuric acid) to nitric acid in the following manner: (9.78) Step I If excess of electrophilic reagent is used, further substitution reaction may take place in which other hydrogen atoms of benzene Step II ring may also be successively replaced by the electrophile. For example, benzene on treatment with excess of chlorine in the presence of anhydrous AlCl3 can be Protonated Nitronium chlorinated to hexachlorobenzene (C6Cl6) nitric acid ion It is interesting to note that in the process of generation of nitronium ion, sulphuric acid serves as an acid and nitric acid as a base. Thus, it is a simple acid-base equilibrium. (b) F o r m a t i o n o f C a r b o c a t i o n (arenium ion): Attack of electrophile results in the formation of σ-complex or (9.79) 3 arenium ion in which one of the carbon is sp Mechanism of electrophilic substitution hybridised. reactions: According to experimental evidences, SE (S = substitution; E = electrophilic) reactions are supposed to proceed via the following three steps: (a) Generation of the eletrophile sigma complex (arenium ion) (b) Formation of carbocation intermediate The arenium ion gets stabilised by resonance:(c) Removal of proton from the carbocation intermediate (a) Generation of electrophile E ⊕: During chlorination, alkylation and acylation of benzene, anhydrous AlCl3, being a Lewis acid helps in generation of the elctrophile Cl⊕, R ⊕, RC⊕O (acylium ion) respectively by combining with the attacking reagent. Reprint 2025-26 324 chemistry Sigma complex or arenium ion loses its chemical equation: aromatic character because delocalisation of 3 CxHy + (x + ) O2 → x CO2 + H2O n (9.83)electrons stops at sp hybridised carbon. (c) Removal of proton: To restore the 9.5.6 Directive influence of a functional aromatic character, σ -complex releases group in monosubstituted benzene proton from sp3 hybridised carbon on attack – When monosubstituted benzene is subjected by [AlCl4] (in case of halogenation, alkylation – to further substitution, three possible and acylation) and [HSO4] (in case of disubstituted products are not formed in nitration). equal amounts. Two types of behaviour are observed. Either ortho and para products or meta product is predominantly formed. It has also been observed that this behaviour depends on the nature of the substituent already present in the benzene ring and not on the nature of the entering group. This is known as directive influence of substituents. Reasons for ortho/para or meta directive nature of groups are discussed below: Addition reactions Ortho and para directing groups: The Under vigorous conditions, i.e., at high groups which direct the incoming group to temperature and/ or pressure in the presence ortho and para positions are called ortho and of nickel catalyst, hydrogenation of benzene para directing groups. As an example, let us gives cyclohexane. discuss the directive influence of phenolic (–OH) group. Phenol is resonance hybrid of following structures: Cyclohexane (9.80) Under ultra-violet light, three chlorine molecules add to benzene to produce benzene hexachloride, C6H6Cl6 which is also called gammaxane. Benzene hexachloride, It is clear from the above resonating (BHC) structures that the electron density is more on (9.81) o – and p – positions. Hence, the substitution Combustion: When heated in air, benzene takes place mainly at these positions. However, burns with sooty flame producing CO2 and it may be noted that –I effect of – OH group also H2O operates due to which the electron density on 15 ortho and para positions of the benzene ring C H6 + O2 → 6CO2 +3H2 O is slightly reduced. But the overall electron 6 2 (9.82) density increases at these positions of the General combustion reaction for any ring due to resonance. Therefore, –OH group hydrocarbon may be given by the following activates the benzene ring for the attack by Reprint 2025-26 Hydrocarbons 325 an electrophile. Other examples of activating In this case, the overall electron density groups are –NH2, –NHR, –NHCOCH3, –OCH3, on benzene ring decreases making further –CH3, –C2H5, etc. substitution difficult, therefore these groups are also called ‘deactivating groups’. TheIn the case of aryl halides, halogens are moderately deactivating. Because of their electron density on o – and p – position strong – I effect, overall electron density on is comparatively less than that at meta benzene ring decreases. It makes further position. Hence, the electrophile attacks on substitution difficult. However, due to comparatively electron rich meta position resonance the electron density on o– and resulting in meta substitution. p– positions is greater than that at the 9.6 Carcinogenicity and Toxicitym-position. Hence, they are also o– and p – directing groups. Resonance structures of Benzene and polynuclear hydrocarbons chlorobenzene are given below: containing more than two benzene rings fused together are toxic and said to possess cancer producing (carcinogenic) property. Such polynuclear hydrocarbons are formed on incomplete combustion of organic materials like tobacco, coal and petroleum. They enter into human body and undergo various biochemical reactions and finally damage DNA and cause cancer. Some of the carcinogenic hydrocarbons are given below (see box). Meta directing group: The groups which direct the incoming group to meta position are called meta directing groups. Some examples of meta directing groups are –NO2, –CN, –CHO, –COR, –COOH, –COOR, –SO3H, etc. Let us take the example of nitro group. Nitro group reduces the electron density in the benzene ring due to its strong–I effect. Nitrobenzene is a resonance hybrid of the following structures. Reprint 2025-26 326 chemistry SUMMARY Hydrocarbons are the compounds of carbon and hydrogen only. Hydrocarbons are mainly obtained from coal and petroleum, which are the major sources of energy. Petrochemicals are the prominent starting materials used for the manufacture of a large number of commercially important products. LPG (liquefied petroleum gas) and CNG (compressed natural gas), the main sources of energy for domestic fuels and the automobile industry, are obtained from petroleum. Hydrocarbons are classified as open chain saturated (alkanes) and unsaturated (alkenes and alkynes), cyclic (alicyclic) and aromatic, according to their structure. The important reactions of alkanes are free radical substitution, combustion, oxidation and aromatization. Alkenes and alkynes undergo addition reactions, which are mainly electrophilic additions. Aromatic hydrocarbons, despite having unsaturation, undergo mainly electrophilic substitution reactions. These undergo addition reactions only under special conditions. Alkanes show conformational isomerism due to free rotation along the C–C sigma bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation is more stable as hydrogen atoms are farthest apart. Alkenes exhibit geometrical (cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond. Benzene and benzenoid compounds show aromatic character. Aromaticity, the property of being aromatic is possessed by compounds having specific electronic structure characterised by Hückel (4n+2)π electron rule. The nature of groups or substituents attached to benzene ring is responsible for activation or deactivation of the benzene ring towards further electrophilic substitution and also for orientation of the incoming group. Some of the polynuclear hydrocarbons having fused benzene ring system have carcinogenic property. EXERCISES 9.1 How do you account for the formation of ethane during chlorination of methane ? 9.2 Write IUPAC names of the following compounds : (a) CH3CH=C(CH3)2 (b) CH2=CH-C≡C-CH3 (c) (d) –CH2–CH2–CH=CH2 (f) CH3(CH2)4 CH (CH2)3 CH3 (e) CH2 –CH (CH3)2 (g) CH3 – CH = CH – CH2 – CH = CH – CH – CH2 – CH = CH2 | C2H5 9.3 For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (a) C4H8 (one double bond) (b) C5H8 (one triple bond) 9.4 Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene Reprint 2025-26 Hydrocarbons 327 9.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’. 9.6 An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’. 9.7 Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene? 9.8 Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene 9.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why? 9.10 Why is benzene extra ordinarily stable though it contains three double bonds? 9.11 What are the necessary conditions for any system to be aromatic? 9.12 Explain why the following systems are not aromatic? (i) (ii) (iii) 9.13 How will you convert benzene into (i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii) p - nitrotoluene (iv) acetophenone? 9.14 In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these. 9.15 What effect does branching of an alkane chain has on its boiling point? 9.16 Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism. 9.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene? 9.18 Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. 9.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? 9.20 How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane 9.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane. 9.22 Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+ (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene (b) Toluene, p-H3C – C6H4 – NO2, p-O2N – C6H4 – NO2. 9.23 Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why? 9.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. 9.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example. Reprint 2025-26

9.5 Aromatic Hydrocarbon These hydrocarbons are also known as ‘arenes’. Since most of them possess pleasant odour (Greek; aroma meaning pleasant smelling), the class of compounds was Methylbenzene 1,2-Dimethylbenzene named as ‘aromatic compounds’. Most of such (Toluene) (o-Xylene)compounds were found to contain benzene ring. Benzene ring is highly unsaturated Reprint 2025-26 Hydrocarbons 319 Friedrich August Kekulé,a German chemist was born in 1829 at Darmsdt in Germany. He became Professor in 1856 and Fellow of Royal Society in 1875. He made major contribution to structural organic chemistry by proposing in 1858 that carbon atoms can join to one another to form chains and later in 1865,he found an answer to the challenging problem of benzene structure by suggesting that these chains can close to form rings. He gave the dynamic structural formula to benzene which forms the basis for its modern electronic structure. He described the discovery of benzene structure later as: FRIEDRICH “I was sitting writing at my textbook,but the work did not progress; my thoughts AUGUST KEKULÉ were elsewhere. I turned my chair to the fire, and dozed. Again the atoms were (7th September gambolling before my eyes. This time the smaller groups kept modestly in the 1829–13th July background. My mental eye, rendered more acute by repeated visions of this 1896) kind, could now distinguish larger structures of manifold conformations; long rows,sometimes more closely fitted together; all twisting and turning in snake like motion. But look! What was that? One of the snakes had seized hold of it’s own tail, and the form whirled mockingly before my eyes. As if by a flash of lightning I woke;.... I spent the rest of the night working out the consequences of the hypothesis. Let us learn to dream, gentlemen, and then perhaps we shall learn the truth but let us beware of making our dreams public before they have been approved by the waking mind.”( 1890). One hundred years later, on the occasion of Kekulé’s centenary celebrations a group of compounds having polybenzenoid structures have been named as Kekulenes. was further found to produce one and only one monosubstituted derivative which indicated that all the six carbon and six hydrogen atoms of benzene are identical. On the basis of this observation August Kekulé in 1865 proposed the following structure for benzene having cyclic arrangement of six carbon atoms with 1,3 Dimethylbenzene 1,4-Dimethylbenzene alternate single and double bonds and one (m-Xylene) ( p-Xylene) hydrogen atom attached to each carbon atom. 9.5.2 Structure of Benzene Benzene was isolated by Michael Faraday in 1825. The molecular formula of benzene, C6H6, indicates a high degree of unsaturation. This molecular formula did not account for its relationship to corresponding alkanes, alkenes and alkynes which you have studied in earlier sections of this unit. What do you The Kekulé structure indicates think about its possible structure? Due to t h e p o s s i b i l i t y o f t w o i s o m e r i c its unique properties and unusual stability, 1, 2-dibromobenzenes. In one of the isomers, it took several years to assign its structure. the bromine atoms are attached to the Benzene was found to be a stable molecule doubly bonded carbon atoms whereas in the and found to form a triozonide which indicates other, they are attached to the singly bonded the presence of three double bonds. Benzene carbons. Reprint 2025-26 320 chemistry unhybridised p orbital perpendicular to the plane of the ring as shown below: However, benzene was found to form only one ortho disubstituted product. This problem was overcome by Kekulé by suggesting the concept of oscillating nature of double bonds in benzene as given below. The unhybridised p orbital of carbon atoms are close enough to form a π bond by lateral overlap. There are two equal possibilities of Even with this modification, Kekulé forming three π bonds by overlap of p orbitalsstructure of benzene fails to explain unusual of C1 –C2, C3 – C4, C5 – C6 or C2 – C3, C4 – C5,stability and preference to substitution C6 – C1 respectively as shown in the followingreactions than addition reactions, which figures. could later on be explained by resonance. Resonance and stability of benzene According to Valence Bond Theory, the concept of oscillating double bonds in benzene is now explained by resonance. Benzene is a hybrid of various resonating structures. The two structures, A and B given by Kekulé are the main contributing structures. The hybrid structure is represented by inserting a circle or a dotted circle in the hexagon as shown in (C). The circle represents the six electrons which are delocalised between the six carbon Fig. 9.7 (a) atoms of the benzene ring. (A) (B) (C) The orbital overlapping gives us better picture about the structure of benzene. All the six carbon atoms in benzene are sp 2 hybridized. Two sp2 hybrid orbitals of each carbon atom overlap with sp2 hybrid orbitals of adjacent carbon atoms to form six C—C sigma bonds which are in the hexagonal 2 Fig. 9.7 (b) plane. The remaining sp hybrid orbital of each carbon atom overlaps with s orbital Structures shown in Fig. 9.7(a) and (b) of a hydrogen atom to form six C—H sigma correspond to two Kekulé’s structure with bonds. Each carbon atom is now left with one localised π bonds. The internuclear distance Reprint 2025-26 Hydrocarbons 321 between all the carbon atoms in the ring has (i) Planarity been determined by the X-ray diffraction to (ii) Complete delocalisation of the π electrons be the same; there is equal probability for the in the ring p orbital of each carbon atom to overlap with (iii) Presence of (4n + 2) π electrons in the ringthe p orbitals of adjacent carbon atoms [Fig. where n is an integer (n = 0, 1, 2, . . .).

6.22 What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN? Answers to Some Intext Questions 6.1 6.2 (i) H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I2. 6.3 (i) ClCH2CH2CH2Cl (ii) ClCH2CHClCH3 (iii) Cl2CHCH2CH3 (iv) CH3CCl2CH3 191 Haloalkanes and Haloarenes Reprint 2025-26

6.9 Which compound in each of the following pairs will react faster in SN2 reaction with –OH? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl 6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane. 6.11 How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Toluene to benzyl alcohol (v) Propene to propyne (vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene to but-2-ene (ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl. 6.12 Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halides, though polar, are immiscible with water? (iii) Grignard reagents should be prepared under anhydrous conditions? 6.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform. 6.14 Write the structure of the major organic product in each of the following reactions: (i) CH3CH2CH2Cl + NaI (ii) (CH3)3CBr + KOH (iii) CH3CH(Br)CH2CH3 + NaOH (iv) CH3CH2Br + KCN (v) C6H5ONa + C2H5Cl (vi) CH3CH2CH2OH + SOCl2 (vii) CH3CH2CH = CH2 + HBr (viii) CH3CH = C(CH3)2 + HBr 6.15 Write the mechanism of the following reaction: nBuBr + KCN nBuCN 6.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane. 6.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH. 6.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss. 6.19 How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-bromopropane Chemistry 190 Reprint 2025-26 (iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide

6.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

6.20 The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

6.4 All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product. The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product. Similarly the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible. 6.5 6.6 (i) Chloromethane, Bromomethane, Dibromomethane, Bromoform. Boiling point increases with increase in molecular mass. (ii) Isopropylchloride, 1-Chloropropane, 1-Chlorobutane. Isopropylchloride being branched has lower b.p. than 1- Chloropropane. 6.7 (i) CH3CH2CH2CH2Br Being primary halide, there won’t be any steric hindrance. (ii) Secondary halide reacts faster than tertiary halide. (iii) The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate. 6.8 (i) Tertiary halide reacts faster than secondary halide because of the greater stability of tert-carbocation. Because of greater stability of secondary carbocation than (ii) primary. 6.9 Chemistry 192 Reprint 2025-26 UnitUnitUnitUnit Unit77 AlcoholsAlcoholsAlcohols,AlcoholsAlcoholsAlcoholsAlcoholsAlcohols,AlcoholsAlcohols PhenolsPhenolsPhenolsPhenolsPhenolsPhenolsPhenolsPhenolsPhenolsPhenolsObjectives After studying this Unit, you will be able to andandandandandandandandandand EtherEthertherthertherstherthertherthersther • name alcohols, phenols and ethers according to the IUPAC system of nomenclature; Alcohols, phenols and ethers are the basic compounds for the • discuss the reactions involved in formation of detergents, antiseptics and fragrances, respectively. the preparation of alcohols from alkenes, aldehydes, ketones and carboxylic acids; You have learnt that substitution of one or more • discuss the reactions involved in hydrogen atom(s) from a hydrocarbon by another atom the preparation of phenols from or a group of atoms result in the formation of an entirely haloarenes, benzene sulphonic new compound having altogether different properties acids, diazonium salts and and applications. Alcohols and phenols are formed cumene; when a hydrogen atom in a hydrocarbon, aliphatic and • discuss the reactions for aromatic respectively, is replaced by –OH group. These preparation of ethers from classes of compounds find wide applications in industry (i) alcohols and (ii) alkyl halides as well as in day-to-day life. For instance, have you and sodium alkoxides/aryloxides; ever noticed that ordinary spirit used for polishing • correlate physical properties of wooden furniture is chiefly a compound containing alcohols, phenols and ethers with their structures; hydroxyl group, ethanol. The sugar we eat, the cotton used for fabrics, the paper we use for writing, are all• discuss chemical reactions of the three classes of compounds on made up of compounds containing –OH groups. Just the basis of their functional think of life without paper; no note-books, books, news- groups. papers, currency notes, cheques, certificates, etc. The magazines carrying beautiful photographs and interesting stories would disappear from our life. It would have been really a different world. An alcohol contains one or more hydroxyl (OH) group(s) directly attached to carbon atom(s), of an aliphatic system (CH3OH) while a phenol contains –OH group(s) directly attached to carbon atom(s) of an aromatic system (C6H5OH). The substitution of a hydrogen atom in a hydrocarbon by an alkoxy or aryloxy group (R–O/Ar–O) yields another class of compounds known as ‘ethers’, for example, CH3OCH3 (dimethyl ether). You may also visualise ethers as compounds formed by Reprint 2025-26 substituting the hydrogen atom of hydroxyl group of an alcohol or phenol by an alkyl or aryl group. In this unit, we shall discuss the chemistry of three classes of compounds, namely — alcohols, phenols and ethers. 7.17.17.17.17.1 ClassificationClassificationClassificationClassificationClassification The classification of compounds makes their study systematic and hence simpler. Therefore, let us first learn how are alcohols, phenols and ethers classified? 7.1.1 Alcohols— Alcohols and phenols may be classified as mono–, di–, tri- or Mono, Di, polyhydric compounds depending on whether they contain one, two, Tri or three or many hydroxyl groups respectively in their structures as Polyhydric given below: alcohols Monohydric Dihydric Trihydric Monohydric alcohols may be further classified according to the hybridisation of the carbon atom to which the hydroxyl group is attached. 3  OH bond: In this class of alcohols, (i) Compounds containing Csp the –OH group is attached to an sp3 hybridised carbon atom of an alkyl group. They are further classified as follows: Primary, secondary and tertiary alcohols: In these three types of alcohols, the –OH group is attached to primary, secondary and tertiary carbon atom, respectively as depicted below: Allylic alcohols: In these alcohols, the —OH group is attached to a sp3 hybridised carbon adjacent to the carbon-carbon double bond, that is to an allylic carbon. For example Benzylic alcohols: In these alcohols, the —OH group is attached to a sp 3—hybridised carbon atom next to an aromatic ring. For example. Chemistry 194 Reprint 2025-26 Allylic and benzylic alcohols may be primary, secondary or tertiary. 2  OH bond: These alcohols contain (ii) Compounds containing Csp —OH group bonded to a carbon-carbon double bond, i.e., to a vinylic carbon or to an aryl carbon. These alcohols are also known as vinylic alcohols. Vinylic alcohol: CH2 = CH – OH 7.1.2 Phenols— Mono, Di and trihydric phenols Monohydric 7.1.3 Ethers Ethers are classified as simple or symmetrical, if the alkyl or aryl groups attached to the oxygen atom are the same, and mixed or unsymmetrical, if the two groups are different. Diethyl ether, C2H5OC2H5, is a symmetrical ether whereas C2H5OCH3 and C2H5OC6H5 are unsymmetrical ethers. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 7.1 Classify the following as primary, secondary and tertiary alcohols: CH3 (i) CH3 C CH2OH (ii) H2C CH CH2OH CH3 OH CH CH3 (iii) CH3 CH2 CH2 OH (iv) CH3 CH CH C OH CH2 CH CH3 (vi) (v) CH3 OH 7.2 Identify allylic alcohols in the above examples. 7.27.27.27.27.2 NomenclatureNomenclatureNomenclatureNomenclatureNomenclature (a) Alcohols: The common name of an alcohol is derived from the common name of the alkyl group and adding the word alcohol to it. For example, CH3OH is methyl alcohol. 195 Alcohols, Phenols and Ethers Reprint 2025-26 According to IUPAC system, the name of an alcohol is derived from the name of the alkane from which the alcohol is derived, by substituting ‘e’ of alkane with the suffix ‘ol’. The position of substituents are indicated by numerals. For this, the longest carbon chain (parent chain) is numbered starting at the end nearest to the hydroxyl group. The positions of the –OH group and other substituents are indicated by using the numbers of carbon atoms to which these are attached. For naming polyhydric alcohols, the ‘e’ of alkane is retained and the ending ‘ol’ is added. The number of –OH groups is indicated by adding the multiplicative prefix, di, tri, etc., before ‘ol’. The positions of –OH groups are indicated by appropriate locants, e.g., HO–CH2–CH2–OH is named as ethane–1, 2-diol. Table 7.1 gives common and IUPAC names of a few alcohols as examples. Table 7.1: Common and IUPAC Names of Some Alcohols Compound Common name IUPAC name CH3 – OH Methyl alcohol Methanol CH3 – CH2 – CH2 – OH n-Propyl alcohol Propan-1-ol Isopropyl alcohol Propan-2-ol CH3 – CH2 – CH2 – CH2 – OH n-Butyl alcohol Butan-1-ol sec-Butyl alcohol Butan-2-ol Isobutyl alcohol 2-Methylpropan-1-ol tert-Butyl alcohol 2-Methylpropan-2-ol HO–H2C–CH2–OH Ethylene glycol Ethane-1,2-diol Glycerol Propane -1, 2, 3-triol Cyclic alcohols are named using the prefix cyclo and considering the —OH group attached to C–1. OH OH CH3 Cyclohexanol 2-Methylcyclopentanol (b) Phenols: The simplest hydroxy derivative of benzene is phenol. It is its common name and also an accepted IUPAC name. As structure of phenol involves a benzene ring, in its substituted compounds the terms ortho (1,2- disubstituted), meta (1,3-disubstituted) and para (1,4-disubstituted) are often used in the common names. Chemistry 196 Reprint 2025-26 OH CH3 CH3 CH3 OH OH OH Common name Phenol o-Cresol m-Cresol p-Cresol IUPAC name Phenol 2-Methylphenol 3-Methylphenol 4-Methylphenol Dihydroxy derivatives of benzene are known as 1, 2-, 1, 3- and 1, 4-benzenediol. OH OH OH OH OH OH Common name Catechol Resorcinol Hydroquinone or quinol IUPAC name Benzene-1,2- diol Benzene- 1,3-diol Benzene- 1,4-diol (c) Ethers: Common names of ethers are derived from the names of alkyl/ aryl groups written as separate words in alphabetical order and adding the word ‘ether’ at the end. For example, CH3OC2H5 is ethylmethyl ether. Table 7.2: Common and IUPAC Names of Some Ethers Compound Common name IUPAC name CH3OCH3 Dimethyl ether Methoxymethane C2H5OC2H5 Diethyl ether Ethoxyethane CH3OCH2CH2CH3 Methyl n-propyl ether 1-Methoxypropane C6H5OCH3 Methyl phenyl ether Methoxybenzene (Anisole) (Anisole) C6H5OCH2CH3 Ethyl phenyl ether Ethoxybenzene (Phenetole) C6H5O(CH2)6 – CH3 Heptyl phenyl ether 1-Phenoxyheptane CH3O CH CH3 Methyl isopropyl ether 2-Methoxypropane CH3 Phenyl isopentyl ether 3- Methylbutoxybenzene CH3– O – CH2 – CH2 – OCH3 — 1,2-Dimethoxyethane — 2-Ethoxy- -1,1-dimethylcyclohexane 197 Alcohols, Phenols and Ethers Reprint 2025-26 If both the alkyl groups are the same, the prefix ‘di’ is added before the alkyl group. For example, C2H5OC2H5 is diethyl ether. According to IUPAC system of nomenclature, ethers are regarded as hydrocarbon derivatives in which a hydrogen atom is replaced by an –OR or –OAr group, where R and Ar represent alkyl and aryl groups, respectively. The larger (R) group is chosen as the parent hydrocarbon. The names of a few ethers are given as examples in Table 7.2. ExampleExampleExampleExampleExample 7.17.17.17.17.1 Give IUPAC names of the following compounds: (i) CH3 CH CH CH CH2OH (ii) CH3 CH O CH2CH3 Cl CH3 CH3 CH3 OH NO2 (iii) H3C CH3 (iv) OC2 H5 SolutionSolutionSolutionSolutionSolution (i) 4-Chloro-2,3-dimethylpentan-1-ol (ii) 2-Ethoxypropane (iii) 2,6-Dimethylphenol (iv) 1-Ethoxy-2-nitrocyclohexane IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 7.3 Name the following compounds according to IUPAC system. (i) (ii) (iii) (iv) (v) 7.37.37.37.37.3 StructuresStructuresStructuresStructuresStructures ofofofofof In alcohols, the oxygen of the –OH group is attached to carbon by a FunctionalFunctionalFunctionalFunctionalFunctional sigma (s ) bond formed3 by the overlap of a sp 3 hybridised orbital of carbon with a sp hybridised orbital of oxygen. Fig. 7.1 depicts GroupsGroupsGroupsGroupsGroups structural aspects of methanol, phenol and methoxymethane. Fig. 7.1: Structures of methanol, phenol and methoxymethane Chemistry 198 Reprint 2025-26 The bond angle in alcohols is slightly less than the tetrahedral angle (109°-28¢). It is due to the repulsion between the unshared electron pairs of oxygen. In phenols, the –OH group is attached to sp2 hybridised carbon of an aromatic ring. The carbon– oxygen bond length (136 pm) in phenol is slightly less than that in methanol. This is due to (i) partial double bond character on account of the conjugation of unshared electron pair of oxygen with the aromatic ring (Section 7.4.4) and (ii) sp 2 hybridised state of carbon to which oxygen is attached. In ethers, the four electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement. The bond angle is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky (–R) groups. The C–O bond length (141 pm) is almost the same as in alcohols. 7.47.47.47.47.4 AlcoholsAlcoholsAlcoholsAlcoholsAlcohols andandandandand 7.4.1 Preparation of Alcohols PhenolsPhenolsPhenolsPhenolsPhenols Alcohols are prepared by the following methods: 1. From alkenes (i) By acid catalysed hydration: Alkenes react with water in the presence of acid as catalyst to form alcohols. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markovnikov’s rule. Mechanism The mechanism of the reaction involves the following three steps: Step 1: Protonation of alkene to form carbocation by electrophilic attack of H3O +. H2O + H+ ® H3O+ Step 2: Nucleophilic attack of water on carbocation. Step 3: Deprotonation to form an alcohol. 199 Alcohols, Phenols and Ethers Reprint 2025-26 Hydroboration - (ii) By hydroboration–oxidation: Diborane (BH3)2 reacts with alkenes oxidation was first to give trialkyl boranes as addition product. This is oxidised to reported by H.C. alcohol by hydrogen peroxide in the presence of aqueous sodium Brown in 1959. For hydroxide.his studies on boron containing organic compounds, Brown shared the 1979 Nobel prize in Chemistry with G. Wittig. The addition of borane to the double bond takes place in such a manner that the boron atom gets attached to the sp2 carbon carrying greater number of hydrogen atoms. The alcohol so formed looks as if it has been formed by the addition of water to the alkene in a way opposite to the Markovnikov’s rule. In this reaction, alcohol is obtained in excellent yield. 2. From carbonyl compounds (i) By reduction of aldehydes and ketones: Aldehydes and ketones are reduced to the corresponding alcohols by addition of hydrogen in the presence of catalysts (catalytic hydrogenation). The usual catalyst is a finely divided metal such as platinum, palladium or nickel. It is also prepared by treating aldehydes and ketones with sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4). Aldehydes yield primary alcohols whereas ketones give secondary alcohols. The numbers in front (ii) By reduction of carboxylic acids and esters: Carboxylic acids of the reagents along are reduced to primary alcohols in excellent yields by lithium the arrow indicate aluminium hydride, a strong reducing agent. that the second reagent is added only (i) LiAlH4 when the reaction RCOOH RCH2 OH with first is complete. (ii) H2O However, LiAlH4 is an expensive reagent, and therefore, used for preparing special chemicals only. Commercially, acids are reduced to alcohols by converting them to the esters (Section 7.4.4), followed by their reduction using hydrogen in the presence of catalyst (catalytic hydrogenation). R'OH H + Chemistry 200 Reprint 2025-26 3. From Grignard reagents Alcohols are produced by the reaction of Grignard reagents (Unit 6, Class XII) with aldehydes and ketones. The first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol. ... (i) ...(ii) The reaction of The overall reactions using different aldehydes and ketones are as Grignard reagents follows: with methanal produces a primary alcohol, with other aldehydes, secondary alcohols and with ketones, tertiary alcohols. You will notice that the reaction produces a primary alcohol with methanal, a secondary alcohol with other aldehydes and tertiary alcohol with ketones. Give the structures and IUPAC names of the products expected from ExampleExampleExampleExampleExample 7.27.27.27.27.2 the following reactions: (a) Catalytic reduction of butanal. (b) Hydration of propene in the presence of dilute sulphuric acid. (c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis. SolutionSolutionSolutionSolutionSolution (a) (b) (c) 7.4.2 Preparation Phenol, also known as carbolic acid, was first isolated in the early of Phenols nineteenth century from coal tar. Nowadays, phenol is commercially produced synthetically. In the laboratory, phenols are prepared from benzene derivatives by any of the following methods: 201 Alcohols, Phenols and Ethers Reprint 2025-26 1. From haloarenes Chlorobenzene is fused with NaOH at 623K and 320 atmospheric pressure. Phenol is obtained by acidification of sodium phenoxide so produced (Unit 6, Class XII). 2. From benzenesulphonic acid Benzene is sulphonated with oleum and benzene sulphonic acid so formed is converted to sodium phenoxide on heating with molten sodium hydroxide. Acidification of the sodium salt gives phenol. 3. From diazonium salts A diazonium salt is formed by treating an aromatic primary amine with nitrous acid (NaNO2 + HCl) at 273-278 K. Diazonium salts are hydrolysed to phenols by warming with water or by treating with dilute acids (Unit 9, Class XII). + – NH2 N2 Cl OH NaNO2 H2O + N2 + HCl +HCl Warm Aniline Benzene diazonium chloride Most of the worldwide 4. From cumene production of phenol is from cumene. Phenol is manufactured from the hydrocarbon, cumene. Cumene (isopropylbenzene) is oxidised in the presence of air to cumene hydroperoxide. It is converted to phenol and acetone by treating it with dilute acid. Acetone, a by-product of this reaction, is also obtained in large quantities by this method. Chemistry 202 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 7.4 Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ? 7.5 Write structures of the products of the following reactions: (i) (ii) (iii) 7.4.3 Physical Alcohols and phenols consist of two parts, an alkyl/aryl group and a Properties hydroxyl group. The properties of alcohols and phenols are chiefly due to the hydroxyl group. The nature of alkyl and aryl groups simply modify these properties. Boiling Points The boiling points of alcohols and phenols increase with increase in the number of carbon atoms (increase in van der Waals forces). In alcohols, the boiling points decrease with increase of branching in carbon chain (because of decrease in van der Waals forces with decrease in surface area). The –OH group in alcohols and phenols is involved in intermolecular hydrogen bonding as shown below: It is interesting to note that boiling points of alcohols and phenols are higher in comparison to other classes of compounds, namely hydrocarbons, ethers, haloalkanes and haloarenes of comparable molecular masses. For example, ethanol and propane have comparable molecular masses but their boiling points differ widely. The boiling point of methoxymethane is intermediate of the two boiling points. 203 Alcohols, Phenols and Ethers Reprint 2025-26 The high boiling points of alcohols are mainly due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons. Solubility Solubility of alcohols and phenols in water is due to their ability to form hydrogen bonds with water molecules as shown. The solubility decreases with increase in size of alkyl/aryl (hydro- phobic) groups. Several of the lower molecular mass alcohols are miscible with water in all proportions. ExampleExampleExampleExampleExample 7.37.37.37.37.3 Arrange the following sets of compounds in order of their increasing boiling points: (a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol. (b) Pentan-1-ol, n-butane, pentanal, ethoxyethane. SolutionSolutionSolutionSolutionSolution (a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol. (b) n-Butane, ethoxyethane, pentanal and pentan-1-ol. 7.4.4 Chemical Alcohols are versatile compounds. They react both as nucleophiles and Reactions electrophiles. The bond between O–H is broken when alcohols react as nucleophiles. Alcohols as nucleophiles (i) (ii) The bond between C–O is broken when they react as electrophiles. Protonated alcohols react in this manner. Protonated alcohols as electrophiles Based on the cleavage of O–H and C–O bonds, the reactions of alcohols and phenols may be divided into two groups: Chemistry 204 Reprint 2025-26 (a) Reactions involving cleavage of O–H bond 1. Acidity of alcohols and phenols (i) Reaction with metals: Alcohols and phenols react with active metals such as sodium, potassium and aluminium to yield corresponding alkoxides/phenoxides and hydrogen. In addition to this, phenols react with aqueous sodium hydroxide to form sodium phenoxides. OH ONa + NaOH + H 2 O Sodium phenoxide The above reactions show that alcohols and phenols are acidic in nature. In fact, alcohols and phenols are Brönsted acids i.e., they can donate a proton to a stronger base (B:). (ii) Acidity of alcohols: The acidic character of alcohols is due to the polar nature of O–H bond. An electron-releasing group (–CH3, –C2H5) increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid strength. For this reason, the acid strength of alcohols decreases in the following order: 205 Alcohols, Phenols and Ethers Reprint 2025-26 Alcohols are, however, weaker acids than water. This can be illustrated by the reaction of water with an alkoxide. This reaction shows that water is a better proton donor (i.e., stronger acid) than alcohol. Also, in the above reaction, we note that an alkoxide ion is a better proton acceptor than hydroxide ion, which suggests that alkoxides are stronger bases (sodium ethoxide is a stronger base than sodium hydroxide). Alcohols act as Bronsted bases as well. It is due to the presence of unshared electron pairs on oxygen, which makes them proton acceptors. (iii) Acidity of phenols: The reactions of phenol with metals (e.g., sodium, aluminium) and sodium hydroxide indicate its acidic nature. The hydroxyl group, in phenol is directly attached to the sp2 hybridised carbon of benzene ring which acts as an electron withdrawing group. Due to this, the charge distribution in phenol molecule, as depicted in its resonance structures, causes the oxygen of –OH group to be positive. The reaction of phenol with aqueous sodium hydroxide indicates that phenols are stronger acids than alcohols and water. Let us examine how a compound in which hydroxyl group attached to an aromatic ring is more acidic than the one in which hydroxyl group is attached to an alkyl group. The ionisation of an alcohol and a phenol takes place as follows: Due to the higher electronegativity of sp2 hybridised carbon of phenol to which –OH is attached, electron density decreases on oxygen. This increases the polarity of O–H bond and results in an increase in ionisation of phenols than that of alcohols. Now let us examine the stabilities of alkoxide and phenoxide ions. In alkoxide ion, the negative charge is localised on oxygen while in phenoxide ion, the charge is delocalised. The delocalisation of negative charge (structures I-V) makes Chemistry 206 Reprint 2025-26 phenoxide ion more stable and favours the ionisation of phenol. Although there is also charge delocalisation in phenol, its resonance structures have charge separation due to which the phenol molecule is less stable than phenoxide ion. In substituted phenols, the presence of electron withdrawing groups such as nitro group, enhances the acidic strength of phenol. This effect is more pronounced when such a group is present at ortho and para positions. It is due to the effective delocalisation of negative charge in phenoxide ion when substituent is at ortho or para position. On the other hand, electron releasing groups, such as alkyl groups, in general, do not favour the formation of phenoxide ion resulting in decrease in acid strength. Cresols, for example, are less acidic than phenol. The greater the pKa Table 7.3: pKa Values of some Phenols and Ethanol value, the weaker the acid. Compound Formula pKa o-Nitrophenol o–O2N–C6H4–OH 7.2 m-Nitrophenol m–O2N–C6H4–OH 8.3 p-Nitrophenol p-O2N–C6H4–OH 7.1 Phenol C6H5–OH 10.0 o-Cresol o-CH3–C6H4–OH 10.2 m-Cresol m-CH3C6H4–OH 10.1 p-Cresol p-CH3–C6H4–OH 10.2 Ethanol C2H5OH 15.9 From the above data, you will note that phenol is million times more acidic than ethanol. Arrange the following compounds in increasing order of their acid strength: ExampleExampleExampleExampleExample 7.47.47.47.47.4 Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol. Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3,5-dinitrophenol, SolutionSolutionSolutionSolutionSolution 2,4, 6-trinitrophenol. 2. Esterification Alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters. 207 Alcohols, Phenols and Ethers Reprint 2025-26 Pyridine R/Ar OH+R’COCl R/ArOCOR’ + HCl The reaction with carboxylic acid and acid anhydride is carried Aspirin possesses out in the presence of a small amount of concentrated sulphuric analgesic, anti- acid. The reaction is reversible, and therefore, water is removed as inflammatory and soon as it is formed. The reaction with acid chloride is carried out in antipyretic properties. the presence of a base (pyridine) so as to neutralise HCl which is formed during the reaction. It shifts the equilibrium to the right hand side. The introduction of acetyl (CH3CO) group in alcohols or phenols is known as acetylation. Acetylation of salicylic acid produces aspirin. (b) Reactions involving cleavage of carbon – oxygen (C–O) bond in alcohols The reactions involving cleavage of C–O bond take place only in alcohols. Phenols show this type of reaction only with zinc. 1. Reaction with hydrogen halides: Alcohols react with hydrogen halides to form alkyl halides (Refer Unit 6, Class XII). ROH + HX ® R–X + H2O The difference in reactivity of three classes of alcohols with HCl distinguishes them from one another (Lucas test). Alcohols are soluble in Lucas reagent (conc. HCl and ZnCl2) while their halides are immiscible and produce turbidity in solution. In case of tertiary alcohols, turbidity is produced immediately as they form the halides easily. Primary alcohols do not produce turbidity at room temperature. 2. Reaction with phosphorus trihalides: Alcohols are converted to alkyl bromides by reaction with phosphorus tribromide (Refer Unit 6, Class XII). 3. Dehydration: Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid e.g., concentrated H2SO4 or H3PO4, or catalysts such as anhydrous zinc chloride or alumina. Ethanol undergoes dehydration by heating it with concentrated H2SO4 at 443 K. Chemistry 208 Reprint 2025-26 Secondary and tertiary alcohols are dehydrated under milder conditions. For example Thus, the relative ease of dehydration of alcohols follows the following order: Tertiary > Secondary > Primary The mechanism of dehydration of ethanol involves the following steps: Tertiary carbocations Mechanism are more stable and Step 1: Formation of protonated alcohol. therefore are easier to form than secondary and primary carbocations; tertiary alcohols are the easiest to dehydrate. Step 2: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction. Step 3: Formation of ethene by elimination of a proton. The acid used in step 1 is released in step 3. To drive the equilibrium to the right, ethene is removed as it is formed. 4. Oxidation: Oxidation of alcohols involves the formation of a carbon- oxygen double bond with cleavage of an O-H and C-H bonds. Such a cleavage and formation of bonds occur in oxidation reactions. These are also known as dehydrogenation reactions as these involve loss of dihydrogen from an alcohol molecule. Depending on the oxidising agent used, a primary alcohol is oxidised to an aldehyde which in turn is oxidised to a carboxylic acid. 209 Alcohols, Phenols and Ethers Reprint 2025-26 Strong oxidising agents such as acidified potassium permanganate are used for getting carboxylic acids from alcohols directly. CrO3 in anhydrous medium is used as the oxidising agent for the isolation of aldehydes. CrO3 RCH 2 OH  RCHO A better reagent for oxidation of primary alcohols to aldehydes in good yield is pyridinium chlorochromate (PCC), a complex of chromium trioxide with pyridine and HCl. PCC CH 3  CH  CH  CH 2 O H  CH 3  CH  CH  CH O Secondary alcohols are oxidised to ketones by chromic anhyride (CrO3). Tertiary alcohols do not undergo oxidation reaction. Under strong reaction conditions such as strong oxidising agents (KMnO4) and elevated temperatures, cleavage of various C-C bonds takes place and a mixture of carboxylic acids containing lesser number of carbon atoms is formed. When the vapours of a primary or a secondary alcohol are passed over heated copper at 573 K, dehydrogenation takes place and an aldehyde or a ketone is formed while tertiary alcohols undergo dehydration. Biological oxidation of methanol and ethanol in the body produces the corresponding aldehyde followed by the acid. At times the alcoholics, by mistake, drink ethanol, mixed with methanol also called denatured alcohol. In the body, methanol is oxidised first to methanal and then to methanoic acid, which may cause blindness and death. A methanol poisoned patient is treated by giving intravenous infusions of diluted ethanol. The enzyme responsible for oxidation of aldehyde (HCHO) to acid is swamped allowing time for kidneys to excrete methanol. (c) Reactions of phenols Following reactions are shown by phenols only. Chemistry 210 Reprint 2025-26 1. Electrophilic aromatic substitution In phenols, the reactions that take place on the aromatic ring are electrophilic substitution reactions (Unit 9, Class XI). The –OH group attached to the benzene ring activates it towards electrophilic substitution. Also, it directs the incoming group to ortho and para positions in the ring as these positions become electron rich due to the resonance effect caused by –OH group. The resonance structures are shown under acidity of phenols. Common electrophilic aromatic substitution reactions taking place in phenol are as follows: (i) Nitration: With dilute nitric acid at low temperature (298 K), phenol yields a mixture of ortho and para nitrophenols. The ortho and para isomers can be separated by steam distillation. o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding which causes the association of molecules. 2, 4, 6 - Trinitrophenol is a strong acid due to the presence of three With concentrated nitric acid, phenol is converted to electron withdrawing 2,4,6-trinitrophenol. The product is commonly known as picric –NO2 groups which acid. The yield of the reaction product is poor. facilitate the release of hydrogen ion. Nowadays picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2,4-disulphonic acid, and then with concentrated nitric acid to get 2,4,6-trinitrophenol. Can you write the equations of the reactions involved? 211 Alcohols, Phenols and Ethers Reprint 2025-26 (ii) Halogenation: On treating phenol with bromine, different reaction products are formed under different experimental conditions. (a) When the reaction is carried out in solvents of low polarity such as CHCl3 or CS2 and at low temperature, monobromophenols are formed. The usual halogenation of benzene takes place in the presence of a Lewis acid, such as FeBr3 (Unit 6, Class XII), which polarises the halogen molecule. In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring. (b) When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate. ExampleExampleExampleExampleExample 7.57.57.57.57.5 Write the structures of the major products expected from the following reactions: (a) Mononitration of 3-methylphenol (b) Dinitration of 3-methylphenol (c) Mononitration of phenyl methanoate. SolutionSolutionSolutionSolutionSolution The combined influence of –OH and –CH3 groups determine the position of the incoming group. 2. Kolbe’s reaction Phenoxide ion generated by treating phenol with sodium hydroxide is even more reactive than phenol towards electrophilic aromatic substitution. Hence, it undergoes electrophilic substitution with carbon dioxide, a weak electrophile. Ortho hydroxybenzoic acid is formed as the main reaction product. Chemistry 212 Reprint 2025-26 3. Reimer-Tiemann reaction On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring. This reaction is known as Reimer - Tiemann reaction. The intermediate substituted benzal chloride is hydrolysed in the presence of alkali to produce salicylaldehyde. 4. Reaction of phenol with zinc dust Phenol is converted to benzene on heating with zinc dust. 5. Oxidation Oxidation of phenol with chromic acid produces a conjugated diketone known as benzoquinone. In the presence of air, phenols are slowly oxidised to dark coloured mixtures containing quinones. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions